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

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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>.
 +
 +
~CHECKMATE2021
  
 
==Solution 2 ~ More efficient for proofs==
 
==Solution 2 ~ More efficient for proofs==
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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>
 
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>
 +
 +
~CHECKMATE2021
  
 
==Solution 3==
 
==Solution 3==
−
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 <math>3</math>. Now, if we just list the first few multiples of 9 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>.
+
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 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
+
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
  
 
==Video Solution (CREATIVE THINKING!!!)==
 
==Video Solution (CREATIVE THINKING!!!)==

Latest revision as of 21:10, 21 January 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 ~ 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 $a=\boxed { \textbf{(E) }7}$

~CHECKMATE2021

Solution 3

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 denominator 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 (CREATIVE THINKING!!!)

https://youtu.be/8WvqSwSG7EQ

~Education, the Study of Everything


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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