2016 AMC 8 Problems/Problem 5

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 $11$ ?

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

Solutions

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 $2$ 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, $73-11(6)=73-66=\boxed{\textbf{(E) }7}$ .

Solution 2 ~ More efficient for proofs

This two digit number must take the form of $10x+y,$ where $x$ and $y$ are integers $0$ to $9.$ However, if x is an integer, we must have $y=3.$ So, the number's new form is $10x+3.$ This needs to have a remainder of $1$ when divided by $9.$ Because of the $9$ divisibility rule, we have $10x+3 \equiv 1 \pmod 9.$ We subtract the three, getting $10x \equiv -2 \pmod 9.$ which simplifies to $10x \equiv 7 \pmod 9.$ However, $9x \equiv 0 \pmod 9,$ so $10x - 9x \equiv 7 - 0 \pmod 9$ and $x \equiv 7 \pmod 9.$

Let the quotient of $9$ in our modular equation be $c,$ and let our desired number be $z,$ so $x=9c+7$ and $z = 10x+3.$ We substitute these values into $z = 10x+3,$ and get $z = 10(9c+7) + 3$ so $z = 90c+73.$ As a result, $z \equiv 73 \pmod {90}.$

Alternatively, we could have also used a system of modular equations to immediately receive $z \equiv 73 \pmod {90}.$

To prove generalization vigorously, we can let $a$ be the remainder when $z$ is divided by $11.$ Setting up a modular equation, we have $90c + 73 \equiv a \pmod {11}.$ Simplifying, $90c+7 \equiv a \pmod {11}$ If $c = 1,$ then we don't have a 2 digit number! Thus, $c=0$ and $a=\boxed { \textbf{(E) }7}$

Video Solution

https://youtu.be/7an5wU9Q5hk?t=574

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.

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