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.
