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 | + | 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== | ||
| + | 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 13:52, 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?
Solution 1
Looking at the values, we notice that 
, 
and 
. This means we are looking for a value that is four less than a multiple of 
, 
, and 
. The least common multiple of these numbers is 
, so the numbers that fulfill this can be written as 
, where 
is a positive integer. This value is only a three digit≤ integer when is 
or 
, which gives 
and 
respectively. Thus we have values, so our answer is 
Solution 2
Let us create the equations: 
, and we know $100 ≤ 11z+7 <1000$ (Error compiling LaTeX. Unknown error_msg), it gives us $9 ≤ z ≤90$ (Error compiling LaTeX. Unknown error_msg), which is the range of the value of z. Because of 
, then 
, so (z+1) must be mutiples of 6. Because of 
, then 
, so (z+1) must also be mutiples of 9. Hence, the value of (z+1) must be common multiples of and , which means multiples of $18(LCM of 6 & 9)$ (Error compiling LaTeX. Unknown error_msg). So let's say 
, then $9 ≤ z = 18p-1 ≤ 90, 1 ≤ p ≤ 91/18 or 1 ≤ p ≤ 5$ (Error compiling LaTeX. Unknown error_msg). Thus our answer is ~LarryFlora
Video Solution
https://youtu.be/CPQpkpnEuIc - Happytwin
https://youtu.be/7an5wU9Q5hk?t=939
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
