Let n be the least positive integer greater than 2021 such that

(n+2)/3,(3n+4)/5,(4n+5)/9

are all integers. Find the sum of n's digits.

Derekshackk Jun 23, 2022

#1**+3 **

We can see that n+2 is divisible by 3, so the lowest possible n that fits this is 2023. We know that 3n+4 is divisible by 5, so n has to end with 2 or 7. The lowest possible value for both statements to hold true is 2032 and we know that 4n+5 is divisible by 9, so the sum of 4n+5's digits have to be divisble by 9 as well, so the sum of 4n's digits have to be 5 less than a multiple of 9. Without the other two statements, the lowest value of this is 2026. You will realize that every single time you add 9 to this number, the property still holds that 4n+5 is divisible by 9. Same goes for the number 2032. So through trial and error we get the lowest number of n is

\(\boxed{2062}\)

SparklingWater2 Jun 23, 2022