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 $11$ ?

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

Video Solution

Solution

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 $2$ in order for it to have the desired remainder $\pmod {10}.$ 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, $73-11(6)=73-66=\boxed{\textbf{(E) }7}$ .

Video Solution

https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5)

2016 AMC 8 (Problems • Answer Key • Resources)
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.

Page

Toolbox

Search

2016 AMC 8 Problems/Problem 5

Video Solution

Solution

Video Solution