Two-digit numbers that are seven times their digit sum

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Find every two-digit number that is exactly seven times the sum of its own digits. For instance, 21 = 7 × (2 + 1). Then: which two-digit numbers are four times their digit sum? Which are equal to their digit sum?

Accepted Answer

21, 42, 63 and 84. Set it up. Let the number be 10a + b, where a is the tens digit (1 to 9) and b the units digit (0 to 9). The condition is 10a + b = 7(a + b), so 10a + b = 7a + 7b, which simplifies to 3a = 6b, that is, a = 2b. List them. b = 1 gives a = 2: 21. b = 2 gives 42. b = 3 gives 63. b = 4 gives 84. b = 5 would need a = 10, not a digit, and b = 0 gives a = 0, not a two-digit number. So exactly four numbers, and they are the multiples of 21. Four times the digit sum. 10a + b = 4(a + b) gives 6a = 3b, so b = 2a: 12, 24, 36, 48. Equal to the digit sum. 10a + b = a + b gives 9a = 0, imp…

Two-digit numbers that are seven times their digit sum

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