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Difference between revisions of "2016 AMC 8 Problems/Problem 5"

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The number <math>72+1=73</math> satisfies both conditions. We subtract the biggest multiple of <math>11</math> less than <math>73</math> to get the remainder. Thus, <math>73-11(6)=73-66=\boxed{\textbf{(E) }7}</math>.
 
The number <math>72+1=73</math> satisfies both conditions. We subtract the biggest multiple of <math>11</math> less than <math>73</math> to get the remainder. Thus, <math>73-11(6)=73-66=\boxed{\textbf{(E) }7}</math>.
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==Video Solution==
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https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5)
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{{AMC8 box|year=2016|num-b=4|num-a=6}}
 
{{AMC8 box|year=2016|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:06, 5 July 2020

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


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
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All AJHSME/AMC 8 Problems and Solutions

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