Difference between revisions of "2018 AMC 8 Problems/Problem 21"
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https://youtu.be/7an5wU9Q5hk?t=939 - pi_is_3.14 | https://youtu.be/7an5wU9Q5hk?t=939 - pi_is_3.14 | ||
− | https://youtu.be/ | + | https://youtu.be/hoCdk8AC-0c |
~savannahsolver | ~savannahsolver |
Revision as of 11:22, 24 February 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
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.