Difference between revisions of "2018 AMC 8 Problems/Problem 21"
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==Solution 2== | ==Solution 2== | ||
− | Let us create the equations: <math>6x+2 = 9y+5 = 11z+7</math>, and we know <math>100 | + | 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 mutiples of 6. Because of <math>9y+5=11z+7</math>, then <math>9y=11z+2=9z+2(z+1)</math>, so (z+1) must also be mutiples of 9. Hence, the value of (z+1) must be common multiples of <math>6</math> and <math>9</math>, which means multiples of <math>18(LCM of 6 & 9)</math>. So let's say <math>z+1 = 18p</math>, then <math>9 ≤ z = 18p-1 ≤ 90, 1 ≤ p ≤ 91/18 or 1 ≤ p ≤ 5</math>. Thus our answer is <math>\boxed{\textbf{(E) }5}</math> ~LarryFlora |
Revision as of 14:53, 30 May 2021
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 (z+1) must be mutiples of 6. Because of
, then
, so (z+1) must also be mutiples of 9. Hence, the value of (z+1) must be common multiples of
and
, which means multiples of $18(LCM of 6 & 9)$ (Error compiling LaTeX. Unknown error_msg). So let's say
, then $9 ≤ z = 18p-1 ≤ 90, 1 ≤ p ≤ 91/18 or 1 ≤ p ≤ 5$ (Error compiling LaTeX. Unknown error_msg). Thus our answer is
~LarryFlora
Video Solution
https://youtu.be/CPQpkpnEuIc - Happytwin
https://youtu.be/7an5wU9Q5hk?t=939
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.