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The 6-digit number 2X574Y is divisible by 36. How many possible solutions are for X?

My naive approach is to write multiples of 36 and see the patterns, but many multiplies must be written to see which one has 574. I have studied Silverman's Number Theory but these types of questions are never discussed.

What is the general method for solving this question? Or at least for this specific question?

This is from math olympiad for middle school students. There are 4 options : 1,2,3,4.

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    Welcome to Math SE. Note any number being divisible by $36$ means it must be divisible by both $9$ (i.e., the sum of digits is a multiple of $9$) and $4$ (i.e., the last $2$ digits must form a number divisible by $4$). – John Omielan Jan 16 '22 at 20:51
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    Divisibility by $36$ implies divisibility by $4$ and by $9$, so $Y$ must be $0$, $4$, or $8$ and $2+X+5+7+4+Y=18+X+Y$ must be divisible by $9$. This means that $X+Y$ must also be divisible by $9$, since $18+X+Y$ and $18$ are. – Geoffrey Trang Jan 16 '22 at 20:55

1 Answers1

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Since $36=4\times 9$ and $(4,9)=1$ , then given number must divisible by both $4$ and $9$.

If last two digit of a number is divisible by$4$, then the number is divisible by$4$.

$Y=0,4,8$

For divisibility by $9$, the sum of the digits of the number must be a multiple of$9$.

For $Y=0$,

$X+18=9k, k\in N$, $X=0$ or $9$

For $Y=4$,

$X+22=9k, k\in N$, $X=5$

For $Y=8$,

$X+26=9k, k\in N$, $X=1$.

Lion Heart
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