How many three-digit numbers are equal to five times the sum of their digits?

MEMEG0D Oct 30, 2024

#1**0 **

*How many three-digit numbers are equal to five times the sum of their digits?*

100a + 10b + c = 5 • (a + b + c)

100a + 10b + c = 5a + 5b + 5c

95a + 5b – 4c = 0

95a = 4c – 5b

Stipulating that "a" cannot be zero; because, if so, the number would be a two-digit number.

If a = 1 then 95 is too large to equal the most that 4c – 5b can be, namely 36 – 45, i.e., – 9

If a = 2 then 190 is too large to equal the most that 4c – 5b can be, namely 36 – 45, i.e., – 9

If a = 3 then 285 is too large, etc., and so on for the rest of the digits that "a" might equal.

I think this little demonstration shows that there is **no solution**.

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Bosco Oct 31, 2024