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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.

 Jun 23, 2022
 #1
avatar+589 
+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 

\(\boxed{2062}\)

 Jun 23, 2022
 #2
avatar+118118 
0

Hi Sparklingwater2.

Thanks for answering.

 

A small suggestion .. I hope you don't mind.

Blocks like that are very uninviting.

If you spread it out, starting by putting each new sentence on a new line, it will be much easier for people to read and comprehend. :)

 Jun 23, 2022

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