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

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~CHECKMATE2021
 
~CHECKMATE2021
  
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==Solution 2 ~ More efficient for proofs==
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==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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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>
 
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We subtract the three, getting <cmath>10x \equiv -2 \pmod 9.</cmath>
 
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which simplifies to <cmath>10x \equiv 7 \pmod 9.</cmath>
 
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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>
 
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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>
 
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*Alternatively, we could have also used a system of modular equations to immediately receive <math>z \equiv 73 \pmod {90}.</math>
 
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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>
 
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==Solution 3==
 
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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 denominator 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>.
 
 
-fn106068
 
-fn106068
  
 
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
 
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 (CREATIVE THINKING!!!)==
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==Video Solution==
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https://youtu.be/8WvqSwSG7EQ
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https://youtu.be/d-bCEDoZEjg?si=VFLhpgyJ_vHhE7h3
  
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~Education, the Study of Everything
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A solution so simple a 12-year-old made it!
  
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~Elijahman~
  
 
==Video Solution by OmegaLearn==
 
==Video Solution by OmegaLearn==

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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