2016 AMC 8 Problems/Problem 5

Revision as of 13:37, 21 December 2016 by StellarG (talk | contribs) (Solution)

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:

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

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

We can use modular arithmetic to solve this. Firstly, we can form the equations:

$1\equiv N \pmod{9}\\ 3\equiv N \pmod{10}.$

Therefore, $N = 9x + 1$ and $N = 10y + 3$. Since the divisibility rule for $10$ is that the last digit has to be $0$, we can say that a number that has a remainder of $3$ when divided by $10$ ends in $3$.

As the number is $1$ more than a multiple of $9$, the multiple of $9$ ends in $2$. The numbers that are greater than $9$ and end in $2$ are: $12, 22, 32, 42, 52, 62, 72, 82, 92$ and so on. We can see that $72$ is the smallest positive multiple of $9$ that ends in a $2$, so $N$ must equal $72 + 1 = 73$.

Now, we need to find the remainder when $73$ is divided by $11$. The largest multiple of $11$ that is less than $73$ is $66$, so $73 - 66 = 7$ is the remainder when $N$ is divided by $11$.

Our answer is $\boxed{\textbf{(E) }7}$.

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

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