2018 AMC 8 Problems/Problem 21
Contents
[hide]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 |
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