Difference between revisions of "2018 AMC 8 Problems/Problem 21"

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==Solution 2==
 
==Solution 2==
Let us create the equations: <math>6x+2 = 9y+5 = 11z+7</math>, and we know <math>100 \leq 11z+7 <1000</math>, it gives us <math>9 \leq z \leq 90</math>, which is the range of the value of z. Because of <math>6x+2=11z+7</math>, then <math>6x=11z+5=6z+5(z+1)</math>, so <math>(z+1)</math> must be a mutiple of 6. Because of <math>9y+5=11z+7</math>, then <math>9y=11z+2=9z+2(z+1)</math>, so <math>(z+1)</math> must also be a mutiple of <math>9</math>. Hence, the value of <math>(z+1)</math> must be common multiples of <math>6</math> and <math>9</math>, which means multiples of <math>18(LCM \text{ of }\ 6, 9)</math>. So let's say <math>z+1 = 18p</math>, then <math>9 \leq z = 18p-1 \leq 90</math>,  so <math>1 \leq p \leq 91/18\ or \ 1 \leq p \leq 5</math>. Thus the answer is <math>\boxed{\textbf{(E) }5}</math> ~LarryFlora
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Let us create the equations: <math>6x+2 = 9y+5 = 11z+7</math>, and we know <math>100 \leq 11z+7 <1000</math>, it gives us <math>9 \leq z \leq 90</math>, which is the range of the value of z. Because of <math>6x+2=11z+7</math>, then <math>6x=11z+5=6z+5(z+1)</math>, so <math>(z+1)</math> must be a mutiple of 6. Because of <math>9y+5=11z+7</math>, then <math>9y=11z+2=9z+2(z+1)</math>, so <math>(z+1)</math> must also be a mutiple of <math>9</math>. Hence, the value of <math>(z+1)</math> must be a common multiple of <math>6</math> and <math>9</math>, which means multiples of <math>18(LCM \text{ of }\ 6, 9)</math>. So let's say <math>z+1 = 18p</math>, then <math>9 \leq z = 18p-1 \leq 90</math>,  so <math>1 \leq p \leq 91/18\ or \ 1 \leq p \leq 5</math>. Thus the answer is <math>\boxed{\textbf{(E) }5}</math> ~LarryFlora
  
 
==Solution 3==
 
==Solution 3==

Revision as of 19:59, 21 December 2021

Problem

How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

Solution 1

Looking at the values, we notice that $11-7=4$, $9-5=4$ and $6-2=4$. This means we are looking for a value that is four less than a multiple of $11$, $9$, and $6$. The least common multiple of these numbers is $11\cdot3^{2}\cdot2=198$, so the numbers that fulfill this can be written as $198k-4$, where $k$ is a positive integer. This value is only a three digit integer when $k$ is $1, 2, 3, 4$ or $5$, which gives $194, 392, 590, 788,$ and $986$ respectively. Thus we have $5$ values, so our answer is $\boxed{\textbf{(E) }5}$

Solution 2

Let us create the equations: $6x+2 = 9y+5 = 11z+7$, and we know $100 \leq 11z+7 <1000$, it gives us $9 \leq z \leq 90$, which is the range of the value of z. Because of $6x+2=11z+7$, then $6x=11z+5=6z+5(z+1)$, so $(z+1)$ must be a mutiple of 6. Because of $9y+5=11z+7$, then $9y=11z+2=9z+2(z+1)$, so $(z+1)$ must also be a mutiple of $9$. Hence, the value of $(z+1)$ must be a common multiple of $6$ and $9$, which means multiples of $18(LCM \text{ of }\ 6, 9)$. So let's say $z+1 = 18p$, then $9 \leq z = 18p-1 \leq 90$, so $1 \leq p \leq 91/18\ or \ 1 \leq p \leq 5$. Thus the answer is $\boxed{\textbf{(E) }5}$ ~LarryFlora

Solution 3

By the Chinese Remainder Theorem, we have that all solutions are in the form $x=198k+194$ where $k\in \mathbb{Z}.$ Counting the number of values, we get $\boxed{\textbf{(E) }5}.$

~mathboy282

Video Solution

https://youtu.be/CPQpkpnEuIc - Happytwin

https://youtu.be/PTwMDbsz2xI

https://youtu.be/7an5wU9Q5hk?t=939

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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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