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

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==Solution==
 
==Solution==
  
From the second bullet point, we know that the second digit must be <math>3</math>. Because there is a remainder of <math>1</math> when it is divided by <math>9</math>, the multiple of <math>9</math> must end in a <math>2</math>. We now look for this one:
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when x equals y what is 2345432x?
  
<math>9(1)=9\\
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IF U DONT KNOW DIS U ARE FREGING IDIOT
9(2)=18\\
 
9(3)=27\\
 
9(4)=36\\
 
9(5)=45\\
 
9(6)=54\\
 
9(7)=63\\
 
9(8)=72</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>.
 
 
 
{{AMC8 box|year=2016|num-b=4|num-a=6}}
 
{{MAA Notice}}
 

Revision as of 17:19, 12 November 2019

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

when x equals y what is 2345432x?

IF U DONT KNOW DIS U ARE FREGING IDIOT

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