2018 AMC 8 Problems/Problem 21: Difference between revisions

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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 mutiples of 6. Because of $9y+5=11z+7$ , then $9y=11z+2=9z+2(z+1)$ , so (z+1) must also be mutiples of 9. Hence, the value of (z+1) must be common multiples 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

2018 AMC 8 (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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

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