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If $792$ divides the integer $13xy45z$, find the digits $x,y$ and $z$.

I know that i have to use some divisibility test but i am stuck how to use it and solve the above example.

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2 Answers 2

up vote 5 down vote accepted

Note that $792=8\times 9\times 11$.

A number is divisible by $8$ if and only if the last three digits are divisible by $8$, so we need $45z$ to be divisible by $8$. That will give you the value of $z$.

A number is divisible by $9$ if and only if the sum of the digits is divisible by $9$. So you need $1+3+x+y+4+5+z$ to be a multiple of $9$. This means that $4+x+y+z$ must be a multiple of $9$. You will already know the value of $z$, so this gives you information about $x+y$.

A number is a multiple of $11$ if and only if the sum of the odd-place digits minus the sum of the even-place digits is a multiple of $11$. So we need $1-3+x-y+4-5+z$ to be a multiple of $11$. You will already have information about $z$, so this will give you information about $x-y$.

From knowing stuff about $x+y$ and about $x-y$, you should be able to figure out both $x$ and $y$.

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Oh! you are right.Yes this gives the answer of my question! –  Kns Apr 14 '12 at 4:03

Some hints:

  1. $792$ is even, so $13xy45z$ is even; this tells you something about the possibilities for $z$. In fact, it’s divisible by $8$, which means that the three-digit number $45z$ has to be divisible by $8$.

  2. $792$ is divisible by $9$, so $13xy45z$ is as well; do you know the easy divisibility test for $9$?

  3. $792$ is divisible by $11$, so $13xy45z$ is as well; do you know the easy divisibility test for $11$?

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