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

m (Solution)
 
(4 intermediate revisions by one other user not shown)
Line 1: Line 1:
 +
== Problem ==
 +
 
The number <math>N</math> is a two-digit number.
 
The number <math>N</math> is a two-digit number.
  
Line 10: Line 12:
 
<math>\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7</math>
 
<math>\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7</math>
  
==Video Solution==
+
==Solutions==
https://youtu.be/7an5wU9Q5hk?t=574
+
===Solution 1===
 
 
==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> in order for it to have the desired remainder<math>\pmod {10}.</math>. We now look for this one:  
+
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> in order for it to have the desired remainder<math>\pmod {10}.</math> We now look for this one:  
  
 
<math>9(1)=9\\
 
<math>9(1)=9\\
Line 27: Line 27:
  
 
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>.
 +
 +
===Solution 2 ~ More efficient for proofs===
 +
 +
This two digit number must take the form of <math>10x+y,</math> where <math>x</math> and <math>y</math> are integers <math>0</math> to <math>9.</math> However, if x is an integer, we must have <math>y=3.</math> So, the number's new form is <math>10x+3.</math> This needs to have a remainder of <math>1</math> when divided by <math>9.</math> Because of the <math>9</math> divisibility rule, we have <cmath>10x+3 \equiv 1 \pmod 9.</cmath>
 +
We subtract the three, getting <cmath>10x \equiv -2 \pmod 9.</cmath>
 +
which simplifies to <cmath>10x \equiv 7 \pmod 9.</cmath>
 +
However, <math>9x \equiv 0 \pmod 9,</math> so <cmath>10x - 9x \equiv 7 - 0 \pmod 9</cmath> and <cmath>x \equiv 7 \pmod 9.</cmath>
 +
 +
Let the quotient of <math>9</math> in our modular equation be <math>c,</math> and let our desired number be <math>z,</math> so <math>x=9c+7</math> and <math>z = 10x+3.</math>  We substitute these values into <math>z = 10x+3,</math> and get <cmath> z = 10(9c+7) + 3</cmath> so <cmath> z = 90c+73.</cmath> As a result, <math>z \equiv 73 \pmod {90}.</math>
 +
 +
*Alternatively, we could have also used a system of modular equations to immediately receive <math>z \equiv 73 \pmod {90}.</math>
 +
 +
To prove generalization vigorously, we can let <math>a</math> be the remainder when <math>z</math> is divided by <math>11.</math> Setting up a modular equation, we have <cmath>90c + 73 \equiv a \pmod {11}.</cmath> Simplifying, <cmath>90c+7 \equiv a \pmod {11}</cmath> If <math>c = 1,</math> then we don't have a 2 digit number! Thus, <math>c=0</math> and <math>a=\boxed { \textbf{(E) }7}</math>
  
 
==Video Solution==
 
==Video Solution==
 
+
https://youtu.be/7an5wU9Q5hk?t=574
https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5)
 
  
  
 
{{AMC8 box|year=2016|num-b=4|num-a=6}}
 
{{AMC8 box|year=2016|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:52, 16 January 2021

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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS