AMC 12, 2009, question 12

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

answer is 1 positive integer exists that satisfy the requirements. (54)

...I just don't know how you solve

thx

Thinking about this on a more fundamental and how numbers are actually formed:

d₁*100+d₂*10+d₃=6(d₁+d₂+d₃)

Where the variables are the first, second and third digits in the number

This simplifies to:

d₁*94+d₂*4-5d₃=0

If d₁ is an integer above zero, then d₃ would have to be a large number greater than 9, but d₃ has to be an integer between 0 and 9, so d₁=0. We are left with: d₂*4-5d₃=0 , where you can see that letting d₂=5 and d₃=4 would work, meaning that the number 54 works.

You can also rewrite this as: d₃/d₂=5/4, which might make it more clear that 54 is the only number that works, because the fraction can't be reduced, and multiplying the numerator and the denominator by 2, the minimal amount, would give 10/8, and you can't have "10" as a single digit, so 54 is the solution.

ah, i see now.

Thanks again