2016 AMC 8 Problems/Problem 5

Revision as of 14:45, 21 December 2016 by StellarG (talk | contribs) (Solution 1)

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 1

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

$91=9\\ 9(2)=18\\ 9(3)=27\\ 9(4)=36\\ 9(5)=45\\ 9(6)=54\\ 9(7)=63\\ 9(8)=72$

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

2016 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png