Three digit multiple of 9

Number Theory Level 2

Find the smallest positive integer that is divisible by 9, and for which all the digits are even.

The answer is 288.

2 solutions

Tushar Malik
Feb 4, 2015

The number clearly has at least two digits. If the number has two digits, then it is of the form ab¯¯¯¯¯, where a and b are even. It is easy to check that no number of this form is divisible by 9 (we can use the fact that a number is divisible by 9 if and only if the sum of its digits is divisible by 9), so the number must have at least three digits. If the number has three digits, then it is of the form abc¯¯¯¯¯¯¯, where a, b, and c are even. If a=2, then only b=c=8 makes the number divisible by 9, so 288 works. Otherwise, a≥4, and the number is greater than 288, so the smallest such number is 288.

Tushar Malik Isn' this problem same as this one?

Anuj Shikarkhane - 6 years, 4 months ago

Yeah, but this is not the first time a problem has appeared more than once on Brilliant. I've seen $(x-a)(x-b)(x-c).......(x-z)$ 14 times.

Satvik Golechha - 6 years, 4 months ago

Bill Bell
Feb 4, 2015

To be divisible by 9, the sum of the required number's digits must be divisible by 9.

This suggested that I could look at partitions of multiples of 9. There are no partitions of 9 lacking an odd digit. So I started looking at ways of partitioning 18 consisting of only even digits. A leading digit of 2 yields no solutions other than 288.

And then because I've learned to be suspicious of any reasoning I do I engaged brute force:

Three digit multiple of 9

The answer is 288.

2 solutions

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