Consider a 3-digit number of three distinct digits. The digits of the number are rearranged to form a new 3-digit number such that the position of each of the digits is changed. The sum of these two numbers is a 4-digit number whose first two digits are the same as the first digit of the original number and the last digit is 8. Also the sum of the first and last digits of the original number is 8. Find the original and the rearranged number.
Question
Consider a 3-digit number of three distinct digits. The digits of the number are rearranged to form a new 3-digit number such that the position of each of the digits is changed. The sum of these two numbers is a 4-digit number whose first two digits are the same as the first digit of the original number and the last digit is 8. Also the sum of the first and last digits of the original number is 8. Find the original and the rearranged number.
Solution 1
Let's break down the problem step by step:
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Let's denote the original 3-digit number as ABC, where A, B, and C are its digits. The rearranged number will be CAB, BAC, BCA, ACB, or CBA.
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According to the problem, A + C = 8. This means that the possible pairs for A and C are (1,7), (2,6), (3,5), or (4,4). However, the problem states that the digits are distinct, so the pair (4,4) is not possible.
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The sum of ABC and its rearrangement is a 4-digit number, which we'll denote as AAB8. This means that B + B (or 2B) must be equal to 10 or 20 because it must carry over to make AAB8 a 4-digit number. However, since B is a single digit, 2B = 20 is not possible. Therefore, 2B = 10, which means B = 5.
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Now we know
Solution 2
Sure, let's solve this step by step.
Step 1: Let's denote the original 3-digit number as ABC, where A, B, and C are its digits. The rearranged number will be CAB.
Step 2: According to the problem, the sum of these two numbers is a 4-digit number whose first two digits are the same as the first digit of the original number and the last digit is 8. So, the sum is AABC8.
Step 3: We know that the sum of two 3-digit numbers cannot exceed 2000. Therefore, the first digit of the 4-digit number (A) must be 1. So, the sum is 11BC8.
Step 4: Also, the sum of the first and last digits of the original number is 8. So, A + C = 8. Since A is 1, C must be 7.
Step 5: Now, we know that the original number is 1B7 and the rearranged number is 71B.
Step 6: Adding these two numbers, we get 11BC8. Since C is 7, B must be 4 to make the sum 8.
So, the original number is 147 and the rearranged number is 714.
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