2016 AMC 8 Problems/Problem 5: Difference between revisions

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 $11$ ?

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

when x equals y what is 2345432x?

IF U DONT KNOW DIS U ARE FREGING IDIOT

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 ==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:
+when x equals y what is 2345432x?
-<math>9(1)=9\\
+IF U DONT KNOW DIS U ARE FREGING IDIOT
-(2)=18\\
-(3)=27\\
-(4)=36\\
-(5)=45\\
-(6)=54\\
-(7)=63\\
-(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}}