2016 AMC 8 Problems/Problem 5

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 $11$ ?

$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7$

Solution

From the second bullet, we know that the second digit must be 3. Because there is a remainder of 1 when it is divided by three, the multiple of 9 must end in a 2. We now look for this one:

$9(1)=9\\ 9(2)=18\\ 9(3)=27\\ 9(4)=36\\ 9(5)=45\\ 9(6)=54\\ 9(7)=63\\ 9(8)=72$

The number $72+1=73$ satisfies both conditions. We subtract the biggest multiple of 11 less than 73 to get the remainder. Thus, $73-11(6)=73-66=\boxed{\text{(E) }7}$ .

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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All AJHSME/AMC 8 Problems and Solutions

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2016 AMC 8 Problems/Problem 5

Solution