2018 AMC 8 Problems/Problem 21
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
Solution 4
We can use modular arithmetic. Set up the equations:
and
These equations can also be written as
and
Since
is congruent to numbers
and
then it must also be congruent to their LCM. Thus,
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
~ethancui0529
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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