32 times a two-digit number equals 23 times the number obtained by reversing…

2026

32 times a two-digit number equals 23 times the number obtained by reversing its digits. The sum of its digits is 15. Find the number.

  1. A.

    70

  2. B.

    69

  3. C.

    96

  4. D.

    87

Attempted by 12 students.

Show answer & explanation

Correct answer: B

For a two-digit number, write it as 10x + y, where x is the tens digit and y is the units digit; the number formed by reversing its digits is then 10y + x. Any relationship stated between a two-digit number and its reverse (a ratio, sum, or difference) becomes a pair of linear equations in x and y that can be solved together.

  1. Let the two-digit number be 10x + y, so its reverse is 10y + x.

  2. The digit-sum condition gives x + y = 15.

  3. "32 times the number equals 23 times its reverse" gives 32(10x + y) = 23(10y + x).

  4. Expanding both sides: 320x + 32y = 230y + 23x.

  5. Rearranging: 320x - 23x = 230y - 32y, i.e. 297x = 198y.

  6. Dividing both sides by 99: 3x = 2y, so y = 1.5x.

  7. Substituting into x + y = 15: x + 1.5x = 15, so 2.5x = 15, giving x = 6.

  8. Then y = 15 - 6 = 9.

  9. So the number is 10x + y = 10(6) + 9 = 69.

Check: the reverse of 69 is 96. 32 x 69 = 2208 and 23 x 96 = 2208, so both sides match, confirming the ratio condition; and 6 + 9 = 15 confirms the digit-sum condition. So 69 satisfies both constraints.

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