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Answer the questions independently of each other.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

Answer the questions independently of each other.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.

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Solution

Sure, let's solve this step by 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 denoted as CAB.

  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 AA8.

  3. Also, the sum of the first and last digits of the original number is 8. So, A + C = 8.

  4. Now, let's add the original number and the rearranged number:

    ABC

  • CAB

AA8

  1. From this, we can see that B + A = 10 (since the sum of two digits cannot exceed 18 and the second digit of the sum is A, B + A must be 10). So, B = 10 - A.

  2. Also, A + C = 10 (since the sum of two digits cannot exceed 18 and the third digit of the sum is 8, A + C must be 10). So, C = 10 - A.

  3. Since A + C = 8, we can substitute C = 10 - A into this equation to get A + 10 - A = 8, which simplifies to A = 2.

  4. Substituting A = 2 into the equations B = 10 - A and C = 10 - A, we get B = 8 and C = 6.

So, the original number is 286 and the rearranged number is 628.

This problem has been solved

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