Question

How many positive integers less than 1000 are 6 times the sum of their digits ?.

Answer 1

Only 1, 54.

There are obviously no such 1-digit numbers (1 - 9).

Let n = ab be a 2-digit number whose value is 6 times its digit sum, meaning

n = 10a + b = 6 (a + b)

Then

10a + b = 6a + 6b

4a = 5b

Now, 4 and 5 are coprime, so for equality to hold, a must be a multiple of 5 and b must be a multiple of 4.

The only valid multiple of 5 is 5 itself:

• a = 5   ⇒   5b = 20   ⇒   b = 4

There are 2 multiples of 4 to check. The first one gives the same result as a = 5.

• b = 8   ⇒   4a = 40   ⇒   a = 10

but a must be 1-digit.

So among the 2-digit numbers, only 1 such number exists, n = 54.

Let n = abc be a 3-digit number satisfying the criterion. Then

n = 100a + 10b + c = 6(a + b + c)

100a + 10b + c = 6a + 6b + 6c

94a + 4b = 5c

c must be a number from 1-9, which means at most 5c = 5•9 = 45.

If we take the smallest possible values of a and b, we have

94•1 + 4•0 = 94

but the smallest value of 94a + 4b is larger than the largest value of 5c.

So there are no such 3-digit numbers.