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

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

Revision as of 14:52, 30 May 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 $100 ≤ 11z+7 <1000$ (Error compiling LaTeX. ! Package inputenc Error: Unicode character ≤ (U+2264)), it gives us $9 ≤ z ≤90$ (Error compiling LaTeX. ! Package inputenc Error: Unicode character ≤ (U+2264)), 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 $18(LCM of 6 & 9)$ (Error compiling LaTeX. ! Misplaced alignment tab character &.). So let's say $z+1 = 18p$, then $9 ≤ z = 18p-1 ≤ 90, 1 ≤ p ≤ 91/18 or 1 ≤ p ≤ 5$ (Error compiling LaTeX. ! Package inputenc Error: Unicode character ≤ (U+2264)). Thus our answer is $\boxed{\textbf{(E) }5}$ ~LarryFlora


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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