Difference between revisions of "2018 AMC 8 Problems/Problem 21"
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==Solution 3== | ==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> | 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> | ||
| + | |||
| + | ==Solution 4== | ||
| + | We can use modular arithmetic. Set up the equations: <math>x &\equiv 2 \mod 6,</math> <math>x &\equiv 5 \mod 9,</math> and <math>x &\equiv 7 \mod 11.</math> These equations can also be written as <math>x+4 &\equiv 0 \mod 6,</math> <math>x+4 &\equiv 0 \mod 9,</math> and <math>x+4 &\equiv 0 \mod 11.</math> Since <math>x+4</math> is congruent to numbers <math>6, 9,</math> and <math>11,</math> then it must also be congruent to their LCM. Thus, <math>x+4 &\equiv 0 \mod 198,</math> since 198 is the LCM of <math>6, 9,</math> and <math>11.</math> Since these numbers have to be three digit, they can only be <math>194, 392, 590, 788,</math> and <math>986.</math> This gives us the answer of <math>\boxed{\textbf{(E) }5}.</math> | ||
~mathboy282 | ~mathboy282 | ||
Revision as of 14:58, 17 December 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 
Solution 4
We can use modular arithmetic. Set up the equations: $x &\equiv 2 \mod 6,$ (Error compiling LaTeX. Unknown error_msg) $x &\equiv 5 \mod 9,$ (Error compiling LaTeX. Unknown error_msg) and $x &\equiv 7 \mod 11.$ (Error compiling LaTeX. Unknown error_msg) These equations can also be written as $x+4 &\equiv 0 \mod 6,$ (Error compiling LaTeX. Unknown error_msg) $x+4 &\equiv 0 \mod 9,$ (Error compiling LaTeX. Unknown error_msg) and $x+4 &\equiv 0 \mod 11.$ (Error compiling LaTeX. Unknown error_msg) Since 
is congruent to numbers 
and 
then it must also be congruent to their LCM. Thus, $x+4 &\equiv 0 \mod 198,$ (Error compiling LaTeX. Unknown error_msg) since 198 is the LCM of and 
Since these numbers have to be three digit, they can only be and 
This gives us the answer of
~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. 
