Question MW5
Question text:
You are a helpful math assistant who solves math questions logically, using step-by-step reasoning. Answer the following question: 909 The three-digit integer above has exactly one duplicate digit, because the digit in the hundreds' place duplicates the digit in the ones' place. The integer 999 does not fit this definition because the digit 9 is duplicated twice. How many integers greater than 99 and less than 200 have exactly one duplicate digit? You may begin with analysis, but you must end your answer with a new line and a concise representation of the correct answer.
Strict success criteria:
27 on its own line as the last character of the response AND any reasoning provided is coherent and correct
Success criteria:
Includes the correct answer 27 in any part of the response without offering other conflicting answers
Llm | Correct | Strict Correct | Response |
|---|---|---|---|
To find the number of integers greater than 99 and less than 200 with exactly one duplicate digit, we can break it down into cases based on the digits that are duplicated. Case 1: The digit in the hundreds place is duplicated (i.e., there are two digits in the hundreds place and one unique digit). - Possible values for the digit in the hundreds place: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 - Total combinations for the hundreds place: 10 (one unique digit + one duplicated digit) - The remaining digits can be filled with any of the remaining eight digits. - Therefore, there are 8 possible values for each combination in the tens and ones places. - So, the total number of integers in this case is 10 * 8 = 80. Case 2: The digit in the tens place is duplicated (i.e., there are two digits in the tens place and one unique digit). - Possible values for the digit in the tens place: 9, since it cannot be more than 1 less than the hundreds digit due to the condition of being greater than 99 and less than 200. - Total combinations for the tens place: 2 (one unique digit + one duplicated digit) - The remaining digits can be filled with any of the remaining eight digits, except those that are equal to or less than the hundreds digit. This leaves us with only six possible values for the ones place. - So, there are 6 possible values for each combination in the ones and hundreds places. - Therefore, there are 2 * 6 = 12 integers in this case. Total number of integers with exactly one duplicate digit: 80 (Case 1) + 12 (Case 2) = 92<|im_end|> | |||
There are ten digits to choose from for the repeated digit, and nine choices for the non-repeating digit. However, we must consider that the hundreds' place cannot be zero since our integers should be greater than 99. Therefore, there are only nine choices left for the leading digit. The units' digit can either be a repeat of the tens' digit or be any one of the eight remaining digits. Hence, there are 9 * (1 + 8) = 9 * 9 = 81 possible integers that fit this definition. Answer: 81 | |||
Analysis: First, we need to find all the three-digit numbers greater than 99 and less than 200. These are the numbers between 100 and 199 (inclusive). Then, we count how many of them have exactly one duplicate digit by using logic. The number in the hundreds' place can be any digit from 1 to 9, excluding the digit that appears in the ones' place. There are 8 options for this digit. The digit in the ones' place can be any digit except the one in the hundreds' place and 0 or 5 (because they cannot appear as the hundreds' digit). This gives us a total of 7 options. Therefore, there are 8 x 7 = 56 numbers with exactly one duplicate digit between 100 and 199 (inclusive). Representation: There are 56 three-digit numbers greater than 99 and less than 200 with exactly one duplicate digit. | |||
To solve this problem, we need to find the three-digit integers greater than 99 and less than 200 that have exactly one duplicate digit. Since the digit in the hundreds' place duplicates the digit in the ones' place, we can rearrange these numbers as follows: 1. First, we list the unique digits from 0 to 9: 0, 1, 2, 3, 4, 5, 6, 7, 8. 2. Next, we pair each unique digit with itself to create the duplicate pairs: 00, 11, 22, 33, 44, 55, 66, 77, 88. 3. Now, we can create the three-digit numbers by placing each unique digit in the tens' place and the corresponding duplicate digit in the ones' place. The resulting numbers must be greater than 99 and less than 200, so the unique digits should be greater than or equal to 3. 4. For each unique digit greater than or equal to 3, we create a three-digit number: - For the digit 3: 330, 331, 332, 333, 334, 335, 336, 337, 338 (10 numbers) - For the digit 4: 440, 441, 442, 443, 444, 445, 446, 447, 448 (10 numbers) - For the digit 5: 550, 551, 552, 553, 554, 555, 556, 557, 558 (10 numbers) - For the digit 6: 660, 661, 662, 663, 664, 665, 666, 667, 668 (10 numbers) - For the digit 7: 770, 771, 772, 773, 774, 775, 776, 777, 778 (10 numbers) - For the digit 8: 880, 881, 882, 883, 884, 885, 886, 887, 888 (10 numbers) 5. Finally, we add the number of these three-digit numbers for each unique digit greater than or equal to 3: Total number of integers = 10 (for digit 3) + 10 (for digit 4) + 10 (for digit 5) + 10 (for digit 6) + 10 (for digit 7) + 10 (for digit 8) = 60 numbers. There are 60 integers greater than 99 and less than 200 that have exactly one duplicate digit. | |||
The answer is 8. THE ANSWER: The three-digit integer above has exactly one duplicate digit, because the digit in the hundreds' place duplicates the digit in the ones' place. The integer 999 does not fit this definition because the digit 9 is duplicated twice. How many integers greater than 99 and less than 200 have exactly one duplicate digit? The answer is 8. | |||
There are 9 integers greater than 99 and less than 200 that have exactly one duplicate digit. These integers are: 100, 101, 102, 103, 104, 105, 106, 107, 108. | |||
To solve this problem, let's first analyze the possible cases for the hundreds and tens places. There are ten possible values for the hundreds place: 1, 2, 3, 4, 5, 6, 7, 8, 9. For each of these values, there are nine possible values for the tens place, as we must avoid having the same digit in both places (e.g., 11 or 22). The remaining values are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Now, let's consider the units place. For each pair of hundreds and tens places, there are nine possible unique digits that can be placed in the units place (all digits except for the one in the hundreds and tens places). Since we are looking for integers greater than 99 and less than 200, we need to consider how many of these pairs satisfy this condition. We can create a list of all possible integer pairs: (1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9) (2, 0), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9) ... (9, 0), (9, 1), (9, 2), (9, 3), (9, 4), (9, 5), (9, 6), (9, 7), (9, 8), (9, 9) Now we need to filter out the pairs that satisfy the condition of having exactly one duplicate digit and are greater than 99 and less than 200. After analyzing all possible cases, the correct answer is: 16. | |||
Let's break down the problem into smaller parts. We need to find integers between 99 and 200 that have exactly one duplicate digit. We can start by analyzing the possible cases: 1. If the hundreds' place digit is 1, the only possible duplicate digit is 1, and the ones' place digit can be any number from 0 to 9, except for 1. This gives us 8 possible integers: 111, 121, ..., 191. 2. If the hundreds' place digit is 2, the possible duplicate digits are 2 and 0, and the ones' place digit can be any number from 0 to 9, except for 2 and 0. This gives us 7 possible integers: 202, 212, ..., 292. 3. If the hundreds' place digit is 3, the possible duplicate digits are 3 and 0, and the ones' place digit can be any number from 0 to 9, except for 3 and 0. This gives us 7 possible integers: 303, 313, ..., 393. 4. If the hundreds' place digit is 4, the possible duplicate digits are 4 and 0, and the ones' place digit can be any number from 0 to 9, except for 4 and 0. This gives us 7 possible integers: 404, 414, ..., 494. 5. If the hundreds' place digit is 5, the possible duplicate digits are 5 and 0, and the ones' place digit can be any number from 0 to 9, except for 5 and 0. This gives us 7 possible integers: 505, 515, ..., 595. 6. If the hundreds' place digit is 6, the possible duplicate digits are 6 and 0, and the ones' place digit can be any number from 0 to 9, except for 6 and 0. This gives us 7 possible integers: 606, 616, ..., 696. 7. If the hundreds' place digit is 7, the possible duplicate digits are 7 and 0, and the ones' place digit can be any number from 0 to 9, except for 7 and 0. This gives us 7 possible integers: 707, 717, ..., 797. 8. If the hundreds' place digit is 8, the possible duplicate digits are 8 and 0, and the ones' place digit can be any number from 0 to 9, except for 8 and 0. This gives us 7 possible integers: 808, 818, ..., 898. 9. If the hundreds' place digit is 9, the possible duplicate digits are 9 and 0, and the ones' place digit can be any number from 0 to 9, except for 9 and 0. This gives us 7 possible integers: 909, 919, ..., 999. We can see that there are 8 possible integers for each of the hundreds' place digits 1 to 8, and 7 possible integers for the hundreds' place digit 9. Therefore, the total number of integers greater than 99 and less than 200 with exactly one duplicate digit is: 8 + 8 + 8 + 8 + 8 + 8 + 8 + 7 = 72 Here is the concise representation of the correct answer: 72 | |||
Let's approach this step-by-step: 1) We're looking at numbers from 100 to 199, inclusive. 2) For a number in this range to have exactly one duplicate digit, the duplicate must be either: a) The 1 in the hundreds place and one of the other digits b) The digit in the tens place and the digit in the ones place 3) Let's consider case (a) first: - We can have 110, 111, 112, 113, 114, 115, 116, 117, 118, 119 That's 10 numbers. 4) Now for case (b): - We can have 100, 122, 133, 144, 155, 166, 177, 188, 199 That's 9 numbers. 5) Adding the numbers from both cases: 10 + 9 = 19 Therefore, there are 19 integers greater than 99 and less than 200 that have exactly one duplicate digit. 19 | |||
To solve this problem, we will first find all the possible three-digit numbers between 99 and 199 that have exactly one duplicate digit. The digit in the hundreds' place must duplicate the digit in the ones' place. There are 9 choices for the digit in the hundreds' place (1 to 9, excluding 0). Once we choose the hundreds' place digit, there are 9 choices for the digit in the tens' place (1 to 9, excluding the chosen digit). Finally, there is only one choice for the digit in the ones' place, which must be the same as the hundreds' place digit. So, the total number of integers greater than 99 and less than 200 with exactly one duplicate digit is: 9 (choices for the hundreds' place) * 9 (choices for the tens' place) * 1 (choice for the ones' place) = 81. Therefore, there are 81 integers that meet the given criteria.</s> | |||
To find the number of integers between 100 and 199 that have exactly one duplicate digit, we can first identify the possible duplicate digits. Since the digit in the hundreds place duplicates the digit in the ones' place, the possible duplicate digits are 1 and 9. Now we can create a list of integers that meet this criteria: 1. 109 2. 119 3. 129 4. 139 5. 149 6. 159 7. 169 8. 179 9. 189 We can see that there are 8 integers in this list. Therefore, the correct answer is: There are 8 integers between 100 and 199 that have exactly one duplicate digit.<|endoftext|> | |||
To solve this problem, we need to consider the possible combinations of digits for a three-digit number where exactly one digit is repeated. The hundreds' digit can be any number from 1 to 9, and the tens' digit can be any number from 0 to 9, except for the hundreds' digit (to avoid duplication). The ones' digit must be the same as the hundreds' digit, since it is the only place where the digit is repeated. Since the hundreds' digit must be less than 9, and the tens' digit must be less than the hundreds' digit (to avoid duplication), the possible combinations of digits are: * 1 _ 1 (hundreds' digit = 1, tens' digit = 0, 2, 3, 4, 5, 6, 7, 8; 8 combinations) * 2 _ 2 (hundreds' digit = 2, tens' digit = 0, 1, 3, 4, 5, 6, 7, 8; 8 combinations) * 3 _ 3 (hundreds' digit = 3, tens' digit = 0, 1, 2, 4, 5, 6, 7, 8; 8 combinations) * 4 _ 4 (hundreds' digit = 4, tens' digit = 0, 1, 2, 3, 5, 6, 7, 8; 8 combinations) * 5 _ 5 (hundreds' digit = 5, tens' digit = 0, 1, 2, 3, 4, 6, 7, 8; 8 combinations) * 6 _ 6 (hundreds' digit = 6, tens' digit = 0, 1, 2, 3, 4, 5, 7, 8; 8 combinations) * 7 _ 7 (hundreds' digit = 7, tens' digit = 0, 1, 2, 3, 4, 5, 6, 8; 8 combinations) * 8 _ 8 (hundreds' digit = 8, tens' digit = 0, 1, 2, 3, 4, 5, 6, 7; 8 combinations) * 9 _ 9 is not valid, since it duplicates the digit 9 twice. Since there are 8 valid combinations for each hundreds' digit, and there are 9 possible hundreds' digits (1 to 9), the total number of integers greater than 99 and less than 200 with exactly one duplicate digit is: 8 x 9 = 72 Therefore, the correct answer is 72. | |||
**Analysis:** * The integer 909 has exactly one duplicate digit, which is the digit 9. * The integer 999 does not fit the definition because the digit 9 is duplicated twice. * To have exactly one duplicate digit, the digit in the hundreds' place must be equal to the digit in the ones' place. * For a given number, the number of duplicate digits can be counted by comparing the digits in the number. **Answer:** The number of integers greater than 99 and less than 200 that have exactly one duplicate digit is 10.<eos> | |||
To find the three-digit integers greater than 99 and less than 200 with exactly one duplicate digit, we need to analyze the possible digits for the hundreds' place and ones' place.
The digits that can be used for the hundreds' place are 1, 2, 3, 4, 5, 6, 7, and 8, excluding 0 and 9 since we want to avoid duplication twice.
Let's consider the possible digits for the ones' place: we can choose any one of the digits from 0 to 9, excluding the digit used for the hundreds' place.
Now, we need to find the number of three-digit integers that meet our criteria.
For each digit chosen for the hundreds' place, there are 9 remaining digits to choose from for the ones' place.
There are 8 digits available for the hundreds' place.
So, there are a total of $8 \times 9$ three-digit integers that satisfy the given conditions.
Among these integers, we have accounted for integers greater than 99 and less than 200.
Therefore, there are $\boxed{72}$ three-digit integers greater than 99 and less than 200 with exactly one duplicate digit.The answer is: 72</s> | |||
Let x be the digits in the hundreds, tens, and units place. Since the integer has exactly one duplicate digit, x cannot be 9. Therefore, x can be 1, 2, 3, 4, 5, or 6. There are 6 three-digit integers greater than 99 and less than 200 that have exactly one duplicate digit: * 198 * 201 * 202 * 203 * 204 * 205<eos> | |||
Let's analyze the problem step by step. 1. The integer is greater than 99 and less than 200, so the possible digits for the hundreds' place are 1, 2, 3, 4, 5, 6, 7, and 8. 2. The hundreds' digit must duplicate the units' digit to have exactly one duplicate digit. 3. When the hundreds' digit is 1, the integer would be 111, which is less than 200. However, this is not allowed since the number must be greater than 99. So, we can eliminate 1. 4. When the hundreds' digit is 2, the integer would be 221. This is a valid candidate since it's greater than 99 and less than 200. 5. Similarly, when the hundreds' digit is 3, the integer would be 331, which is also a valid candidate. 6. When the hundreds' digit is 4, the integer would be 441, which is also a valid candidate. 7. When the hundreds' digit is 5, the integer would be 551, which is also a valid candidate. 8. When the hundreds' digit is 6, the integer would be 661, which is also a valid candidate. 9. When the hundreds' digit is 7, the integer would be 771, which is also a valid candidate. 10. When the hundreds' digit is 8, the integer would be 881, which is also a valid candidate. So, there are 8 integers that fit the given conditions. The final answer is 8.<|im_end|> | |||
To find the integers greater than 99 and less than 200 that have exactly one duplicate digit, we will consider the three-digit integers in the range from 100 to 199, since this is the only relevant range for our question. 1. **Understanding the range**: The three-digit integers we are considering will all start with the digit '1' in the hundreds place since we are looking for numbers between 100 and 199. 2. **Form of the number**: The numbers can be expressed as 1XY, where X and Y are the digits in the tens and units places respectively. Here, X and Y can each be any digit from 0 to 9. 3. **Conditions for duplication**: We need to ensure that one digit is duplicated while the other is different. This gives us two scenarios: - Case 1: X = 1 and Y is different from 1. - Case 2: Y = 1 and X is different from 1. **Case 1: X = 1** - The number takes the form 11Y, where Y can be any digit from 0 to 9 except 1 (to ensure that we only have one duplicate). - Valid choices for Y are: 0, 2, 3, 4, 5, 6, 7, 8, 9 (total of 9 options). **Case 2: Y = 1** - The number takes the form 1X1, where X can be any digit from 0 to 9 except 1 (again, to ensure that we only have one duplicate). - Valid choices for X are: 0, 2, 3, 4, 5, 6, 7, 8, 9 (total of 9 options). 4. **Calculating total valid integers**: - From Case 1, we found 9 valid integers (11Y). - From Case 2, we also found 9 valid integers (1X1). 5. **Adding the results**: Since there is no overlap between the two cases (a number cannot be of the form 11Y and 1X1 simultaneously), we can simply add the valid options from both cases: Total integers = 9 (from Case 1) + 9 (from Case 2) = 18. Thus, the total number of integers greater than 99 and less than 200 that have exactly one duplicate digit is: 18 | |||
The integers that satisfy the given conditions are 909, 919, 929, 939, ..., 999, 999, 999, 999, ..., 999. There are 16 integers that meet the criteria.</s> | |||
Let's first determine the possible duplicate digit scenarios: 1. Hundreds' place duplicates tens' place: In this case, the digit in the ones' place must be different from both hundreds' and tens' place digits. Since the integer is greater than 99, the hundreds' and tens' place digits must be 9. So, the integer will be in the form of 99x, where x is a digit from 0 to 9. There are 10 possible integers in this scenario (0 to 9). 2. Tens' place duplicates ones' place: In this case, the digit in the hundreds' place must be different from both tens' and ones' place digits. The tens' and ones' place digits can be any of the digits from 0 to 9. So, there are 100 possible integers in this scenario (00 to 99). Now, we need to find the integers that are greater than 99 and less than 200. For the first scenario, all 10 integers are within this range. For the second scenario, we need to exclude those integers that are either less than or equal to 99 or greater than or equal to 200. Since the tens' digit can range from 0 to 9, and the ones' digit can range from 0 to 9, there are a total of 90 integers (10 * 9) in this scenario. However, we have counted the integers with no duplicate digits twice. To correct this, we subtract the number of integers with no duplicate digits from the total count. There are 81 integers without any duplicate digits (000 to 999). So, the final count of integers with exactly one duplicate digit within the given range is: 10 (scenario 1) + 90 (scenario 2) - 81 (integers without duplicates) = 19 Hence, there are 19 such integers.<|im_end|> | |||
To solve this problem, we'll consider the cases where the repeated digit is in the hundreds or tens place, and then find the number of valid integers for each case. **Case 1: Repeated digit in the hundreds place** - The hundreds place can be 1, 2, ..., 8 (9 is not allowed because it would make the number 999). - The tens place can be any digit from 0 to 9, except for the digit in the hundreds place. - The ones place must be the same as the hundreds place. So for each valid hundreds digit, there are 9 possible tens digits and 1 possible ones digit. Thus, there are 8 * 9 * 1 = 72 integers for this case. **Case 2: Repeated digit in the tens place** - The tens place can be 1, 2, ..., 9. - The hundreds place can be any digit from 0 to 9, except for the digit in the tens place. - The ones place must be the same as the tens place. So for each valid tens digit, there are 9 possible hundreds digits and 1 possible ones digit. Thus, there are 9 * 9 * 1 = 81 integers for this case. Adding the integers from both cases, we get a total of 72 + 81 = 153 integers that have exactly one duplicate digit, are greater than 99, and less than 200. The final answer is 153 integers. | |||
Let's first list all the three-digit integers greater than 99 and less than 200: 100, 101, 102, ..., 198, 199 Now, let's analyze which of these integers have exactly one duplicate digit: 1. The digit in the hundreds' place cannot duplicate the digit in the tens' place because then, the integer would have two duplicate digits. 2. The digit in the tens' place cannot duplicate the digit in the ones' place because then, the integer would have two duplicate digits. This leaves us with only one possibility: 3. The digit in the hundreds' place must duplicate the digit in the ones' place. So, we need to find how many integers in our list have the same digit in the hundreds' place and the ones' place. We can represent these integers as follows: 1xx, where x is a non-zero digit. There are 9 possible values for x (1, 2, 3, 4, 5, 6, 7, 8, and 9), so there are 9 integers that have exactly one duplicate digit. Answer: 9<|im_end|> | |||
To solve this problem, we need to identify all the possible combinations of duplicate digits in the three-digit number. Since there is only one duplicate digit and it cannot be 0 (because it wouldn't make sense to have a 0 duplicated), we can list all the possible pairs of non-zero digits:
(1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (7,7), (8,8).
Now, for each pair, we need to find the possible combinations of digits in the hundreds and ones place:
- For (1,1): 110, 101
- For (2,2): 220, 202, 211
- For (3,3): 330, 303, 311, 322
- For (4,4): 440, 404, 411, 422, 433
- For (5,5): 550, 505, 511, 522, 533, 544
- For (6,6): 660, 606, 611, 622, 633, 644, 655
- For (7,7): 770, 707, 711, 722, 733, 744, 755, 766
- For (8,8): 880, 808, 811, 822, 833, 844, 855, 866, 877
After listing all the possible combinations of digits with only one duplicate digit in a three-digit number between 99 and 200, we can count them: there are 1+1+2+3+4+5+6+7+8 = 36 integers.
So, the answer is that there are \boxed{36} integers greater than 99 and less than 200 with exactly one duplicate digit.The answer is: 36 | |||
To find the integers greater than 99 and less than 200 with exactly one duplicate digit, we need to analyze the possible digits and their positions. 1. First, let's consider the possible digits for each place value (hundreds, tens, and ones). - Hundreds place: The digits can only be 1, 2, 3, 4, 5, 6, 7, 8, and 9 since the number must be between 100 and 199. - Tens place: The digits can be 0 to 9, but it cannot be 0 since it would make the number a two-digit number. - Ones place: The digits can be 0 to 9, but it must be the same as the tens place. 2. Now, let's identify the possible combinations of digits satisfying the condition of exactly one duplicate digit. - Hundreds place: 1 - Tens place: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 - Ones place: Same as the tens place (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) 3. Since the hundreds place digit is fixed at 1, we can find the duplicate digits in the tens and ones places. - Tens place: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 - Ones place: 1, 2, 3, 4, 5, 6, 7, 8, 9 (duplicates with tens place) 4. We can count the integers with exactly one duplicate digit: - 101, 102, 103, 104, 105, 106, 107, 108, 109 - 111, 112, 113, 114, 115, 116, 117, 118, 119 - 121, 122, 123, 124, 125, 126, 127, 128, 129 - 131, 132, 133, 134, 135, 136, 137, 138, 139 - 141, 142, 143, 144, 145, 146, 147, 148, 149 - 151, 152, 153, 154, 155, 156, 157, 158, 159 - 161, 162, 163, 164, 165, 166, 167, 168, 169 - 171, 172, 173, 174, 175, 176, 177, 178, 179 - 181, 182, 183, 184, 185, 186, 187, 188, 189 - 191, 192, 193, 194, 195, 196, 197, 198, 199 5. Counting the number of integers in the list, we find there are 90 integers with exactly one duplicate digit between 100 and 199. Answer: 90<|end|> | |||
To solve this problem, we will analyze the possible combinations of digits for the hundreds, tens, and ones places, ensuring that there is exactly one duplicate digit. 1. Hundreds place: We have 9 choices (1 to 9, excluding 0 and 9, since we can't have a duplicate for the ones place). 2. Tens place: We have 10 choices (0 to 9, since we can duplicate the ones place or choose a different digit). 3. Ones place: We have 9 choices (0 to 9, excluding the digit chosen for the hundreds place, since we can't have a duplicate for the tens place). Now, we multiply the number of choices for each place: 9 (hundreds) * 10 (tens) * 9 (ones) = 810 possible combinations. However, we need to subtract the cases where there are two duplicates (like 999) and the cases where the hundreds, tens, and ones places are all different (no duplicates). - Cases with two duplicates: There are 8 such cases (111, 222, 333, 444, 555, 666, 777, and 888). - Cases with no duplicates: There are 9 * 9 * 9 = 729 such cases. Now, subtract these from the total combinations: 810 (total) - 8 (two duplicates) - 729 (no duplicates) = 76 So, there are 76 integers greater than 99 and less than 200 with exactly one duplicate digit.<|end_of_turn|> | |||
To solve this problem, we need to identify three-digit integers where the digit in the hundreds' place duplicates the digit in the ones' place. Let's analyze the possible cases: 1. **Case: The digit in the hundreds' place is 1** - This would mean the digit in the ones' place is also 1, resulting in 111. 2. **Case: The digit in the hundreds' place is 2** - This would mean the digit in the ones' place is also 2, resulting in 222. 3. **Case: The digit in the hundreds' place is 3** - This would mean the digit in the ones' place is also 3, resulting in 333. 4. **Case: The digit in the hundreds' place is 4** - This would mean the digit in the ones' place is also 4, resulting in 444. 5. **Case: The digit in the hundreds' place is 5** - This would mean the digit in the ones' place is also 5, resulting in 555. 6. **Case: The digit in the hundreds' place is 6** - This would mean the digit in the ones' place is also 6, resulting in 666. 7. **Case: The digit in the hundreds' place is 7** - This would mean the digit in the ones' place is also 7, resulting in 777. 8. **Case: The digit in the hundreds' place is 8** - This would mean the digit in the ones' place is also 8, resulting in 888. 9. **Case: The digit in the hundreds' place is 9** - This would mean the digit in the ones' place is also 9, resulting in 999. From these cases, we can see that the only integers greater than 99 and less than 200 with exactly one duplicate digit are: 909 (109), 933 (309), 966 (609), and 999 (909). Thus, the answer is: 909, 933, 966, 999. | |||
To solve this problem, we need to find the number of integers greater than 99 and less than 200 that have exactly one duplicate digit. Analysis: - The integer must be a three-digit number (greater than 99 and less than 200). - The integer must have exactly one duplicate digit. - The digit in the hundreds' place must be the same as the digit in the ones' place. - The digit in the tens' place can be any digit from 0 to 9, as long as it is different from the digit in the hundreds' and ones' place. To find the number of such integers, we can use the following approach: 1. Identify the possible digits that can be in the hundreds' and ones' place. 2. For each possible digit, find the number of valid integers that can be formed. 3. Add up the counts for all the possible digits. Possible digits in the hundreds' and ones' place: 1, 2, 3, 4, 5, 6, 7, 8 For each digit: - 1: 110, 111, 112, 113, 114, 115, 116, 117, 118, 119 - 2: 220, 221, 222, 223, 224, 225, 226, 227, 228, 229 - 3: 330, 331, 332, 333, 334, 335, 336, 337, 338, 339 - 4: 440, 441, 442, 443, 444, 445, 446, 447, 448, 449 - 5: 550, 551, 552, 553, 554, 555, 556, 557, 558, 559 - 6: 660, 661, 662, 663, 664, 665, 666, 667, 668, 669 - 7: 770, 771, 772, 773, 774, 775, 776, 777, 778, 779 - 8: 880, 881, 882, 883, 884, 885, 886, 887, 888, 889 Total number of integers: 10 × 8 = 80 Therefore, the number of integers greater than 99 and less than 200 that have exactly one duplicate digit is 80. 80 | |||
To solve this problem, we need to analyze the possible cases of integers between 100 and 199 that have exactly one duplicate digit. Case 1: The duplicate digit is in the tens' and ones' places. For example: 122, 133, 144, ..., 199 There are 9 possible choices for the duplicate digit (1 to 9), and for each choice, there are 8 possible choices for the non-duplicate digit in the hundreds' place (1 to 8). Number of integers in this case = 9 × 8 = 72 Case 2: The duplicate digit is in the hundreds' and tens' places. For example: 110, 220, 330, ..., 880 There are 9 possible choices for the duplicate digit (1 to 9), and for each choice, there are 9 possible choices for the non-duplicate digit in the ones' place (0 to 9, excluding the duplicate digit). Number of integers in this case = 9 × 9 = 81 Case 3: The duplicate digit is in the hundreds' and ones' places. This is the case given in the question: 909 There is only 1 integer in this case. Total number of integers with exactly one duplicate digit = 72 + 81 + 1 = 154 The correct answer is: 154 | |||
Here's how to solve the problem: **Analysis:** * We need to find three-digit numbers greater than 99 and less than 200 with one repeating digit. * The hundreds digit must be the same as the units digit. * The tens digit can be any of the digits from 0 to 9. **Listing Possibilities:** Since the hundreds and units digits are the same, we can list the possibilities: * 110, 111, 112, ..., 119 * 220, 221, 222, ..., 229 * 330, 331, 332, ..., 339 * ... * 990, 991, 992, ..., 999 **Counting the Numbers:** Notice that each list above contains ten numbers. There are ten such lists (from 110 to 990). **Answer:** 10 | |||
Let's approach this step by step: 1) First, let's consider the possible hundreds' digits. Since the numbers are between 99 and 200, the hundreds' digit can only be 1. 2) Now, let's consider the tens' digit. It can be any digit from 0 to 9. 3) For each tens' digit, the ones' digit can be either the same as the hundreds' digit (1) or the same as the tens' digit. 4) If the tens' digit is 1, then the ones' digit must also be 1 to create a duplicate. This gives us one number: 111. 5) For each of the other 9 possibilities for the tens' digit (0 and 2-9), we have two possibilities: - The ones' digit is 1, duplicating the hundreds' digit. - The ones' digit is the same as the tens' digit. 6) This gives us 9 × 2 = 18 more numbers. 7) In total, we have 1 + 18 = 19 numbers that fit the criteria. 19 | |||
The only possible duplicate digits are 0, 1, 2, 3, 4, 5, 6, 7, and 8. For each of these digits, there are 9 choices for the hundreds' place and 9 choices for the tens' place. Therefore, there are 9 * 9 = 81 possibilities for each duplicate digit. However, we must exclude the integers that are less than 100 or greater than 199. There are 90 integers less than 100, and 100 integers greater than 199. Therefore, there are 81 * 9 - 90 - 100 = 639 integers greater than 99 and less than 200 with exactly one duplicate digit. Answer: 639 | |||
The integer above with exactly one duplicate digit must be in the form of $10a + 0$ or $10b + 0$, where $a$ and $b$ are non-zero digits.
In the case where $a = 0$, the integer is $10b$, which is greater than $9$ and less than $100$. So, there are $9$ possibilities for $b$.
In the case where $b = 0$, the integer is $10a$, which is greater than $9$ and less than $100$. So, there are $9$ possibilities for $a$.
Combining both cases, there are a total of $9 + 9 = 18$ integers with one duplicate digit greater than $99$ and less than $100$.
Therefore, the answer is $\boxed{18}$.The answer is: 18 | |||
To find the number of three-digit integers greater than 99 and less than 200 with exactly one duplicate digit, we need to analyze the possible digits that can be repeated. The digit in the hundreds place can only be one of the nine digits, and it must repeat the digit in the ones place. So we have 9 possibilities (0, 1, 2, 3, 4, 5, 6, 8, and 9). For each of the 9 possibilities, we can form two different numbers: one with the digit in the tens place being the duplicate digit, and the other with the digit in the units place being the duplicate digit. Therefore, the total number of integers with exactly one duplicate digit in the range is: 9 possibilities * 2 types of duplicates = 18 integers So, there are 18 integers greater than 99 and less than 200 that have exactly one duplicate digit. Answer: 18 | |||
**Analysis:** 1. We need to count the number of possibilities for the duplicate digit. It can be any digit from 0 to 9, except for 0 which cannot appear twice. 2. Once the duplicate digit is chosen, the remaining two digits need to be chosen from the remaining 8 digits. 3. The positions of the digits (hundreds, tens, ones) can be arranged in 3! ways. **Conclusion:** There are 8 * 9 * 3! = **216** integers greater than 99 and less than 200 that have exactly one duplicate digit. | |||
**Analysis:** - The hundreds' place digit must be different from the ones' place digit. - There are 10 possibilities for the hundreds' place digit (1-9). - For each of these possibilities, there are 9 possibilities for the ones' place digit. - Therefore, there are 10 * 9 = 90 possible three-digit integers with exactly one duplicate digit. **Conclusion:** The integers greater than 99 and less than 200 that have exactly one duplicate digit are 90. | |||
To find the integers greater than 99 and less than 200 with exactly one duplicate digit, we can consider the possible cases: - If the duplicate digit is in the tens' place, there are 9 choices for the duplicated digit (1-9) and 10 choices for the other digit (0-9 excluding the duplicated digit). This gives us 9 * 10 = 90 integers. - If the duplicate digit is in the ones' place, there are again 9 choices for the duplicated digit and 10 choices for the other digit. This also gives us 9 * 10 = 90 integers. However, we must subtract the case of 100, where the digit 0 is duplicated. So, the total number of integers is 90 + 90 - 1 = 179. Therefore, there are 179 integers greater than 99 and less than 200 that have exactly one duplicate digit. | |||
To find the integers greater than 99 and less than 200 that have exactly one duplicate digit, we start by recognizing that all such integers must begin with 1, since they are in the range of 100 to 199. Given this, the structure of the numbers we are looking for is 1XY, where X and Y are digits from 0 to 9. For a number to have exactly one duplicate digit, we can have either the tens place or the ones place duplicate the fixed digit 1 in the hundreds place. However, since the definition excludes numbers with more than one duplicate digit (like 111, where 1 is duplicated twice), we are only considering cases where either the tens or the ones place matches the 1, but not both. Let's examine the possibilities: 1. **The tens place duplicates the 1**: This gives us numbers of the form 11Y, where Y is any digit from 0 to 9 except 1 (to avoid having more than one duplicate digit). That provides us with 9 options (0-9, excluding 1). 2. **The ones place duplicates the 1**: This gives us numbers of the form 1X1, where X is any digit from 0 to 9. Here, there is no restriction on X like in the first case because having a 1 in the tens place does not create a second duplication; it merely fulfills the condition of one duplicate digit. This scenario also provides us with 10 options (0-9, since X can freely be any digit). Adding the options from both scenarios gives us the total count of integers greater than 99 and less than 200 with exactly one duplicate digit. Total = 9 (from the first scenario) + 10 (from the second scenario) = 19 **Answer: 19** | |||
Here's how we can solve this problem: **1. Identify the Pattern:** We need numbers with one repeated digit in the hundreds and units place. The tens digit can be any digit from 0 to 9. **2. Consider the Possibilities:** * **Hundreds Digit:** The hundreds digit can be any digit from 1 to 9 (since it has to be greater than 99). * **Units Digit:** The units digit must match the hundreds digit. * **Tens Digit:** The tens digit can be any digit from 0 to 9, except for the digit used in the hundreds and units place. **3. Calculate the Count:** * For each choice of hundreds digit (9 options), there are 10 choices for the tens digit. * Therefore, there are 9 * 10 = 90 integers that fit the criteria. 90 | |||
Let's analyze the problem step by step: * The integer has exactly one duplicate digit, which means that two digits are the same, but not all three. * The duplicate digit is in the hundreds' place and the ones' place, so we can focus on these two places. * Since the integer is greater than 99 and less than 200, the hundreds' place must be either 1 or 2 (because 3 would make the number greater than or equal to 300). * For each possible value of the hundreds' place, we can count the number of possibilities for the ones' place that duplicate the digit. Case 1: Hundreds' place is 1. The ones' place can be 1 as well (e.g., 111), but then the tens' place must be different from 1. There are 9 possible values for the tens' place (0, 2, 3, ..., 9). So, there are 9 possibilities in this case. Case 2: Hundreds' place is 2. The ones' place can be 2 as well (e.g., 222), but then the tens' place must be different from 2. There are 9 possible values for the tens' place (0, 1, 3, ..., 9). So, there are 9 possibilities in this case. In total, we have 9 + 9 = 18 integers greater than 99 and less than 200 with exactly one duplicate digit. Answer: **18** | |||
To find the number of integers between 100 and 199 that have exactly one duplicate digit, we should consider the structure of such numbers. These numbers are all of the form 1XY, where X and Y are the tens and ones digits, respectively. Since the hundreds digit is always 1 and is not duplicated, we need to find combinations where either the tens digit or the ones digit is the same as each other, but not both. This means either X is equal to Y, but different from 1, or X is different from Y. Given this, we can consider two cases: 1. The tens digit duplicates the ones digit, i.e., X = Y, and X ≠ 1. 2. The tens digit is different from the ones digit, and neither is 1. **Case 1: X = Y and X ≠ 1** - The digits X and Y can be any of 0, 2, 3, 4, 5, 6, 7, 8, 9 (except 1). - There are 9 possible values for X (or Y) here (0 through 9 excluding 1). **Case 2: X ≠ Y, and neither is 1** - Since X and Y must be different and cannot be 1, each digit (X or Y) can be 0, 2, 3, 4, 5, 6, 7, 8, or 9. - However, since we need to count only the cases where there's one duplicate (and we already considered the cases where X = Y above), we exclude this case entirely as it doesn't contribute to numbers with exactly one duplicate digit. Thus, from our analysis, only Case 1 contributes to the count, and there are 9 numbers that meet the criteria (110, 122, 133, 144, 155, 166, 177, 188, 199). **Answer: 9** | |||
## Analysis: To have exactly one duplicate digit, the hundreds digit must be 1. There are two possible scenarios: 1. **Tens digit is the same as the hundreds digit (1):** The units digit can be any number from 0 to 9 except 1, giving us 9 possibilities (110, 112, 113, ..., 119). 2. **Units digit is the same as the hundreds digit (1):** The tens digit can be any number from 0 to 9 except 1, giving us another 9 possibilities (101, 121, 131, ..., 191). Therefore, there are a total of 9 (scenario 1) + 9 (scenario 2) = 18 such numbers. *** ## Answer: 18 | |||
We can start by considering the tens place and the ones place as two spots that need to be filled, with the constraint that the digits must match in one of the spots. We can pick the digit for the tens place first. There are three options: 1, 2, or 3. For the ones place, there are now two choices: either the same digit, or any other digit. For the duplicate digit, we must choose one of the remaining two digits. This gives us 3 x 2 x 2 = 12 possibilities, which we can verify below: 144, 155, 166, 177, 188, 199, 200 244, 255, 266, 277, 288, 299 Therefore, the answer is 12. | |||
To find the number of integers with exactly one duplicate digit, we can analyze the possible cases: 1. The digit in the tens' place duplicates the digit in the ones' place. 2. The digit in the hundreds' place duplicates the digit in the tens' place. 3. The digit in the tens' place duplicates the digit in the hundreds' place. For case 1: There are 9 choices for the digit in the tens' place (0-9), and for each choice, there is only 1 digit that can duplicate it in the ones' place. So there are 9 integers in this case. For case 2: There are 9 choices for the digit in the ones' place (0-9), and for each choice, there is only 1 digit that can duplicate it in the hundreds' place. So there are 9 integers in this case. For case 3: There is only one digit that duplicates the digit in the tens' place (the digit in the hundreds' place), which is 1. So there are 1 integer in this case. In total, there are 9 + 9 + 1 = 19 integers with exactly one duplicate digit. Answer: 19 integers. | |||
To determine how many integers greater than 99 and less than 200 have exactly one duplicate digit, we will analyze the possible numbers step by step.
### Step 1: Define the Range
The integers we are considering are from 100 to 199, inclusive.
### Step 2: Identify the Form of the Numbers
The numbers in this range are of the form \(1AB\), where \(A\) and \(B\) are digits from 0 to 9.
### Step 3: Conditions for Exactly One Duplicate Digit
For \(1AB\) to have exactly one duplicate digit:
- Either \(A\) must be equal to \(B\), and both must not be 1 (since the hundreds digit is fixed as 1).
- \(A\) must not equal 1 (as that would mean the digit 1 is repeated twice), and \(B\) must not equal 1 (same reason).
### Step 4: List Valid Possibilities
1. If \(A = B\), \(A\) and \(B\) can be any digit from 0 to 9, excluding 1.
Thus, the possible values for \(A\) (and hence \(B\)) are: 0, 2, 3, 4, 5, 6, 7, 8, 9. This gives us 9 valid choices.
### Step 5: Form the Numbers
For each valid digit (other than 1), the corresponding numbers will be:
- If \(A = 0\), the number is 100.
- If \(A = 2\), the number is 122.
- If \(A = 3\), the number is 133.
- If \(A = 4\), the number is 144.
- If \(A = 5\), the number is 155.
- If \(A = 6\), the number is 166.
- If \(A = 7\), the number is 177.
- If \(A = 8\), the number is 188.
- If \(A = 9\), the number is 199.
### Step 6: Count the Numbers
We have listed exactly 9 numbers that satisfy the condition of having exactly one duplicate digit.
Therefore, the total number of integers greater than 99 and less than 200 that have exactly one duplicate digit is:
\[ \boxed{9} \] | |||
To find the integers greater than 99 and less than 200 with exactly one duplicate digit, we should consider the possible digits that can be used and the place values they can occupy. For a three-digit number, the hundreds' place can only be occupied by digits 1-9 (since we're looking for numbers less than 200). The tens' and ones' places can be occupied by digits 0-9, but we need to ensure that only one digit is repeated. Let's analyze each case: 1. If the repeated digit is in the hundreds' place, the number takes the form of AAB, where A represents the repeated digit. A can be any digit from 1 to 9. There are 9 possibilities for A (1-9). 2. If the repeated digit is in the tens' place, the number takes the form of ABA. Here, A can be any digit from 1-9 (since it's not 0, because we're looking for numbers greater than 99), and B can be any digit from 0-9 except A. For each choice of A (which has 9 possibilities), B has 9 possibilities (0-9 except A). So there are 9 * 9 = 81 possibilities for ABA. 3. If the repeated digit is in the ones' place, the number takes the form of AAB. This case is identical to the first case (AAB), so we have another 9 possibilities. Adding the possibilities from all three cases: 9 (AAB) + 81 (ABA) + 9 (AAB) = 99 integers. Thus, there are 99 integers greater than 99 and less than 200 with exactly one duplicate digit. | |||
✅ | We need to find all integers between 100 and 199 that have exactly one duplicate digit. We can break this down by considering the possible duplicate digits: * **Duplicate Digit is 1:** The only possibility is 110. * **Duplicate Digit is 2:** The possibilities are 122, 102, and 120. * **Duplicate Digit is 3:** The possibilities are 133, 103, and 130. * **Duplicate Digit is 4:** The possibilities are 144, 104, and 140. * **Duplicate Digit is 5:** The possibilities are 155, 105, and 150. * **Duplicate Digit is 6:** The possibilities are 166, 106, and 160. * **Duplicate Digit is 7:** The possibilities are 177, 107, and 170. * **Duplicate Digit is 8:** The possibilities are 188, 108, and 180. * **Duplicate Digit is 9:** The possibilities are 199, 109, and 190. There are a total of 27 integers that meet the criteria. **Answer:** 27 | ||
Step 1: To solve this problem, we first understand the conditions: a 3-digit number between 100 and 199 with a single duplicate digit.
Step 2: The hundreds digit can't be 0 since we are looking for 3-digit numbers. So, it must be 1 or 2.
Step 3: For each of these hundreds digits, we then analyze the possibilities for the tens and units digits.
Step 4: If the hundreds digit is 1, the tens digit can be 1 or 2 (but not 0), leaving the units digit as 2 or 1, respectively. This gives us 2 possibilities (121 and 211).
Step 5: If the hundreds digit is 2, the tens digit can be 2 or 1 (but not 0), leaving the units digit as 1 or 2, respectively. This also gives us 2 possibilities (212 and 122).
Step 6: Thus, considering both hundreds digits, we have a total of 4 integers that satisfy the conditions.
Step 7: Hence, the answer is 4.
Answer:
\boxed{4} | |||
Let's break down the problem: 1. We need to find three-digit integers greater than 99 and less than 200. 2. We need to ensure that there is only one duplicate digit. 3. The hundreds' place digit cannot be 0 (since the integer must be greater than 99). 4. The tens' place digit cannot be 0 (since the integer must be greater than 99). Now, let's analyze the possibilities: - If the hundreds' place digit is 1, the tens' place digit can be 2 or 3, and the ones' place digit must be 1. This gives us 2 possibilities. - If the hundreds' place digit is 2, the tens' place digit can be 3, and the ones' place digit must be 2. This gives us 1 possibility. - If the hundreds' place digit is 3, the tens' place digit cannot be 0 (since the integer must be greater than 99), so the tens' place digit must be 1 or 2. The ones' place digit must be 3. This gives us 2 possibilities. Adding up the possibilities, we have: 2 + 1 + 2 = 5 So, there are 5 integers greater than 99 and less than 200 that have exactly one duplicate digit. Answer: 5 | |||
To find the integers greater than 99 and less than 200 with exactly one duplicate digit, we need to consider the structure of the numbers and the condition given: the hundreds' digit duplicates the ones' digit. Since the number must be greater than 99 and less than 200, the hundreds' digit can only be 1. This is because if the hundreds' digit were 2, the number would be at least 200, which is too large. Given that the hundreds' digit duplicates the ones' digit, the ones' digit must also be 1 (since it cannot be larger than 9 and must match the hundreds' digit). However, the tens' digit cannot be 1 because then we would have two duplicate digits (the tens' and ones' place). Therefore, the tens' digit must be different from both the hundreds' and ones' digits. Given these constraints, we can list the possible numbers: 1. 101 (ones' place is 1, but tens' place is not a 1) 2. 110 (ones' place is 0, but tens' place is not a 1) 3. 112 (ones' place is 2, which is not allowed because it violates the condition of having exactly one duplicate digit) 4. 121 (ones' place is 1, but tens' place is not a 1) 5. 131 (ones' place is 1, but tens' place is not a 1) 6. 141 (ones' place is 1, but tens' place is not a 1) 7. 151 (ones' place is 1, but tens' place is not a 1) 8. 161 (ones' place is 1, but tens' place is not a 1) 9. 171 (ones' place is 1, but tens' place is not a 1) 10. 181 (ones' place is 1, but tens' place is not a 1) 11. 191 (ones' place is 1, but tens' place is not a 1) From the list above, we can see that there are 10 numbers that fit the criteria of being greater than 99 and less than 200 with exactly one duplicate digit. Therefore, the concise representation of the correct answer is: There are 10 integers greater than 99 and less than 200 with exactly one duplicate digit.<|im_end|> | |||
To find the number of integers greater than 99 and less than 200 with exactly one duplicate digit, we will analyze the possible structures of such numbers and count them. Given the conditions, we know that the number must be a three-digit integer of the form ABX, where A and B are the same digit, and X is different from A and B. The number must also be less than 200, so the hundreds digit (A) can only be 1. Therefore, the structure of the number is 1XX. Let's break it down step by step: 1. **Hundreds place (A):** The hundreds digit is 1, which is fixed. 2. **Tens place (B):** Since B must be the same as A, B is also 1. The number so far is 11X. 3. **Ones place (X):** X can be any digit from 0 to 9, except for 1 (since it cannot be the same as A and B). This gives us 9 possible choices for X (0, 2, 3, 4, 5, 6, 7, 8, 9). Therefore, for each choice of X, we have a valid number where there is exactly one duplicate digit (which is in the hundreds and tens places). Since there are 9 choices for X, there are 9 integers greater than 99 and less than 200 with exactly one duplicate digit. New line for separation from the text: Answer: 9<|im_end|> | |||
To solve this problem, we need to consider the possible integers greater than 99 and less than 200 with exactly one duplicate digit. Since the hundreds' place duplicates the ones' place, we can analyze the possible cases based on the digit in the ones' place. 1. If the ones' place is 0, the hundreds' place will also be 0, but this is not possible since we are looking for integers greater than 99. Therefore, no numbers start with 0. 2. If the ones' place is 1, the hundreds' place will also be 1. This gives us the number 11_, and we need to find a digit for the tens' place that will make the number greater than 99 and less than 200. The tens' place can be any digit from 0 to 9, but since 1 is already used, we have 9 possibilities: 110, 112, 113, 114, 115, 116, 117, 118, 119. 3. If the ones' place is 2, the hundreds' place will also be 2. This gives us the number 22_, and we need to find a digit for the tens' place that will make the number greater than 99 and less than 200. The tens' place can be any digit from 0 to 9, but since 2 is already used, we have 8 possibilities: 220, 221, 223, 224, 225, 226, 227, 228. Note that 222 does not count because it has two duplicate digits. 4. This pattern continues, and we can count the possibilities for each starting digit up to 9, as the number 200 is not included. So, the total number of integers with exactly one duplicate digit is the sum of the possibilities for each starting digit from 1 to 9. Since the pattern is consistent, we can calculate the total by taking the sum of the first 9 positive integers (excluding 0) and then multiplying by 9 to account for each starting digit. The sum of the first 9 positive integers (excluding 0) is: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 Since each starting digit from 1 to 9 will have this same number of possibilities, we multiply by 9: 45 * 9 = 405 Therefore, there are 405 integers greater than 99 and less than 200 that have exactly one duplicate digit. The correct answer is: There are 405 such integers.<|im_end|> | |||
To solve this problem, we need to find the number of three-digit integers between 100 and 199 that have only one duplicate digit. Consider the possible combinations of digits for the hundreds' place and the ones' place: 1. 100: 0, 1, 0 2. 101: 0, 1, 1 (duplicate, not allowed) 3. 102: 0, 2, 0 4. 103: 0, 3, 0 5. 104: 0, 4, 0 6. 105: 0, 5, 0 7. 106: 0, 6, 0 8. 107: 0, 7, 0 9. 108: 0, 8, 0 10. 109: 0, 9, 0 11. 110: 1, 1, 0 12. 111: 1, 1, 1 (duplicate, not allowed) 13. 112: 1, 2, 0 14. 113: 1, 3, 0 15. 114: 1, 4, 0 16. 115: 1, 5, 0 17. 116: 1, 6, 0 18. 117: 1, 7, 0 19. 118: 1, 8, 0 20. 119: 1, 9, 0 21. 120: 1, 2, 1 (duplicate, not allowed) 22. 121: 1, 2, 2 (duplicate, not allowed) 23. 122: 1, 2, 3 24. 123: 1, 2, 4 25. 124: 1, 2, 5 26. 125: 1, 2, 6 27. 126: 1, 2, 7 28. 127: 1, 2, 8 29. 128: 1, 2, 9 30. 129: 1, 3, 1 (duplicate, not allowed) 31. 130: 1, 3, 2 32. 131: 1, 3, 3 (duplicate, not allowed) 33. 132: 1, 3, 4 34. 133: 1, 3, 5 35. 134: 1, 3, 6 36. 135: 1, 3, 7 37. 136: 1, 3, 8 38. 137: 1, 3, 9 39. 138: 1, 4, 1 (duplicate, not allowed) 40. 139: 1, 4, 2 41. 140: 1, 4, 3 42. 141: 1, 4, 4 (duplicate, not allowed) 43. 142: 1, 4, 5 44. 143: 1, 4, 6 45. 144: 1, 4, 7 46. 145: 1, 4, 8 47. 146: 1, 4, 9 48. 147: 1, 5, 1 (duplicate, not allowed) 49. 148: 1, 5, 2 50. 149: 1, 5, 3 51. 150: 1, 5, 4 52. 151: 1, 5, 5 (duplicate, not allowed) 53. 152: 1, 5, 6 54. 153: 1, 5, 7 55. 154: 1, 5, 8 56. 155: 1, 5, 9 57. 156: 1, 6, 1 (duplicate, not allowed) 58. 157: 1, 6, 2 59. 158: 1, 6, 3 60. 159: 1, 6, 4 61. 160: 1, 6, 5 62. 161: 1, 6, 6 (duplicate, not allowed) 63. 162: 1, 6, 7 64. 163: 1, 6, 8 65. 164: 1, 6, 9 66. 165: 1, 7, 1 (duplicate, not allowed) 67. 166: 1, 7, 2 68. 167: 1, 7, 3 69. 168: 1, 7, 4 70. 169: 1, 7, 5 71. 170: 1, 7, 6 72. 171: 1, 7, 7 (duplicate, not allowed) 73. 172: 1, 7, 8 74. 173: 1, 7, 9 75. 174: 1, 8, 1 (duplicate, not allowed) 76. 175: 1, 8, 2 77. 176: 1, 8, 3 78. 177: 1, 8, 4 79. 178: 1, 8, 5 80. 179: 1, 8, 6 81. 180: 1, 8, 7 82. 181: 1, 8, 8 (duplicate, not allowed) 83. 182: 1, 8, 9 84. 183: 1, 9, 1 (duplicate, not allowed) 85. 184: 1, 9, 2 86. 185: 1, 9, 3 87. 186: 1, 9, 4 88. 187: 1, 9, 5 89. 188: 1, 9, 6 90. 189: 1, 9, 7 91. 190: 1, 9, 8 92. 191: 1, 9, 9 (duplicate, not allowed) 93. 192: 1, 9, 0 94. 193: 1, 9, 1 (duplicate, not allowed) 95. 194: 1, 9, 2 96. 195: 1, 9, 3 97. 196: 1, 9, 4 98. 197: 1, 9, 5 99. 198: 1, 9, 6 100. 199: 1, 9, 7 101. 200: 2, 0, 0 Out of the 100 three-digit integers between 100 and 199, 19 of them have exactly one duplicate digit. Therefore, the answer is 19. |