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## Competency in Focus: Divisibility.

This problem from American Mathematics contest (AMC 8, 2016) is based on the concept of divisibility .

## Next understand the problem

The number $N$ is a two-digit number. • When $N$ is divided by $9$, the remainder is $1$. • When $N$ is divided by $10$, the remainder is $3$. What is the remainder when $N$ is divided by $11$?   $\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7$
##### Source of the problem
American Mathematical Contest 2016, AMC 8  Problem 5

### Divisibility

4/10
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Do you really need a hint? Try it first!
When $N$ is divided by $10$ it leaves remainder $3$ i.e., $N=10\times P+3 \textbf{ ,where } P$ is an integer. i.e., The unit digit of $N$ must be $3$ because unit digit of $10P$ is zero.
$N$ leaves remainder $1$ when divided by $9$. i.e., $N=9\times Q+1$ where $Q$ is an integer. Since $10\times P+3=9\times Q+1$ the unit digit of $9\times Q +1$ must be $3$.
Since $N$ is a two digit number then the only possibility is $Q=8$ i.e., $N=9\times 8+1=73$

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