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

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== Problem ==
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The number <math>N</math> is a two-digit number.
 
The number <math>N</math> is a two-digit number.
  
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==Solution 1==
 
==Solution 1==
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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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From the second bullet point, we know that the second digit must be <math>3</math>, for a number divisible by <math>10</math> ends in zero. Since there is a remainder of <math>1</math> when <math>N</math> is divided by <math>9</math>, the multiple of <math>9</math> must end in a <math>2</math> 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\\
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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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~CHECKMATE2021
  
 
==Solution 2==
 
==Solution 2==
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We know that the number has to be one more than a multiple of <math>9</math>, because of the remainder of one, and the number has to be <math>3</math> more than a multiple of <math>10</math>, which means that it has to end in a <math>3</math>. Now, if we just list the first few multiples of <math>9</math> adding one to the number we get: <math>10, 19, 28, 37, 46, 55, 64, 73, 82, 91</math>. As we can see from these numbers,  the only one that has a three in the units place is <math>73</math>, thus we divide <math>73</math> by <math>11</math>, getting <math>6</math> <math>R7</math>, hence, <math>\boxed{\textbf{(E) }7}</math>.
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-fn106068
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We could also remember that, for a two-digit number to be divisible by <math>9</math>, the sum of its digits has to be equal to <math>9</math>. Since the number is one more than a multiple of <math>9</math>, the multiple we are looking for has a ones digit of <math>2</math>, and therefore a tens digit of <math>9-2 = 7</math>, and then we could proceed as above. -vaisri
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==Video Solution==
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https://youtu.be/d-bCEDoZEjg?si=VFLhpgyJ_vHhE7h3
  
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We can use modular arithmetic to solve this.
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A solution so simple a 12-year-old made it!
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Firstly, we can form the equations:
 
  
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<math>1\equiv N \pmod{9}\\
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~Elijahman~
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3\equiv N \pmod{10}.</math>
 
  
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Therefore, <math>N = 9x + 1</math> and <math>N = 10y + 3</math>.
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==Video Solution by OmegaLearn==
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Since the divisibility rule for <math>10</math> is that the last digit has to be <math>0</math>, we can say that a number that has a remainder of <math>3</math> when divided by <math>10</math> ends in <math>3</math>.
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https://youtu.be/7an5wU9Q5hk?t=574
  
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As the number is <math>1</math> more than a multiple of <math>9</math>, the multiple of <math>9</math> ends in <math>2</math>. The numbers that are greater than <math>9</math> and end in <math>2</math> are:
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==Video Solution==
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<math>12, 22, 32, 42, 52, 62, 72, 82, 92</math> and so on. We can see that <math>72</math> is the smallest positive multiple of <math>9</math> that ends in a <math>2</math>, so <math>N</math> must equal <math>72 + 1 = 73</math>.
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https://youtu.be/aKWQl7kEMy0
  
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Now, we need to find the remainder when <math>73</math> is divided by <math>11</math>. The largest multiple of <math>11</math> that is less than <math>73</math> is <math>66</math>, so <math>73 - 66 = 7</math> is the remainder when <math>N</math> is divided by <math>11</math>.
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~savannahsolver
  
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Our answer is <math>\boxed{\textbf{(E) }7}</math>.
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==See Also==
 
{{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 15:52, 1 July 2024

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 $11$?


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

Solution 1

From the second bullet point, we know that the second digit must be $3$, for a number divisible by $10$ ends in zero. Since there is a remainder of $1$ when $N$ is divided by $9$, the multiple of $9$ must end in a $2$ 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, $73-11(6)=73-66=\boxed{\textbf{(E) }7}$.

~CHECKMATE2021

Solution 2

We know that the number has to be one more than a multiple of $9$, because of the remainder of one, and the number has to be $3$ more than a multiple of $10$, which means that it has to end in a $3$. Now, if we just list the first few multiples of $9$ adding one to the number we get: $10, 19, 28, 37, 46, 55, 64, 73, 82, 91$. As we can see from these numbers, the only one that has a three in the units place is $73$, thus we divide $73$ by $11$, getting $6$ $R7$, hence, $\boxed{\textbf{(E) }7}$. -fn106068

We could also remember that, for a two-digit number to be divisible by $9$, the sum of its digits has to be equal to $9$. Since the number is one more than a multiple of $9$, the multiple we are looking for has a ones digit of $2$, and therefore a tens digit of $9-2 = 7$, and then we could proceed as above. -vaisri

Video Solution

https://youtu.be/d-bCEDoZEjg?si=VFLhpgyJ_vHhE7h3

A solution so simple a 12-year-old made it!

~Elijahman~

Video Solution by OmegaLearn

https://youtu.be/7an5wU9Q5hk?t=574

Video Solution

https://youtu.be/aKWQl7kEMy0

~savannahsolver

See Also

2016 AMC 8 (Problems • Answer Key • Resources)
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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