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> |
==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 (z+1) must be | + | 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 |
− | ==Video | + | ==Solution 3== |
+ | By the [[Chinese Remainder Theorem]], we have that all solutions are in the form <math>x=198k+194</math> where <math>k\in \mathbb{Z}.</math> Counting the number of values, we get <math>\boxed{\textbf{(E) }5}.</math> | ||
+ | |||
+ | ~mathboy282 | ||
+ | |||
+ | ==Video Solutions== | ||
https://youtu.be/CPQpkpnEuIc - Happytwin | https://youtu.be/CPQpkpnEuIc - Happytwin | ||
https://youtu.be/PTwMDbsz2xI | https://youtu.be/PTwMDbsz2xI | ||
− | https://youtu.be/7an5wU9Q5hk?t=939 | + | https://youtu.be/7an5wU9Q5hk?t=939 - pi_is_3.14 |
+ | |||
+ | https://youtu.be/hoCdk8AC-0c | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | https://www.youtube.com/watch?v=PjYwbGm_2aM | ||
==See Also== | ==See Also== |
Latest revision as of 17:34, 23 April 2022
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 , it gives us , which is the range of the value of z. Because of , then , so must be a mutiple of 6. Because of , then , so must also be a mutiple of . Hence, the value of must be a common multiple of and , which means multiples of . So let's say , then , so . Thus the answer is ~LarryFlora
Solution 3
By the Chinese Remainder Theorem, we have that all solutions are in the form where Counting the number of values, we get
~mathboy282
Video Solutions
https://youtu.be/CPQpkpnEuIc - Happytwin
https://youtu.be/7an5wU9Q5hk?t=939 - pi_is_3.14
~savannahsolver
https://www.youtube.com/watch?v=PjYwbGm_2aM
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.