Difference between revisions of "2018 AMC 8 Problems/Problem 21"
(→Solution 2) |
Mathboy282 (talk | contribs) (solution 3 - chinese remainder theorem) |
||
| Line 10: | Line 10: | ||
==Solution 2== | ==Solution 2== | ||
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 \text{ of }\ 6, 9)</math>. So let's say <math>z+1 = 18p</math>, then <math>9 \leq z = 18p-1 \leq 90</math>, so <math>1 \leq p \leq 91/18\ or \ 1 \leq p \leq 5</math>. Thus the answer is <math>\boxed{\textbf{(E) }5}</math> ~LarryFlora | 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 \text{ of }\ 6, 9)</math>. So let's say <math>z+1 = 18p</math>, then <math>9 \leq z = 18p-1 \leq 90</math>, so <math>1 \leq p \leq 91/18\ or \ 1 \leq p \leq 5</math>. Thus the answer is <math>\boxed{\textbf{(E) }5}</math> ~LarryFlora | ||
| + | ==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> | ||
| + | |||
| + | ~mathboy282 | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 00:44, 4 December 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 
. 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 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. 
