What is the smallest number which when divided by 8,12 or 15 leaves remainder of 11 ?
Hello, haftakhan!
What is the smallest number which when divided by 8, 12 or 15
leaves a remainder of 11?
skeeter is absolutely right.
I'll assume that the problem is poorly worded.
The expected answer is probably:
. . $\displaystyle \text{LCM}(8,12,15) + 11 \:=\:120 + 11 \:=\:\boxed{131}$
Check
. . $\displaystyle 131 \div 15 \:=\:8,\;r\,11$
. . $\displaystyle 131 \div 12 \:=\:10,\;r\,11$
And I suppose we can write:
. . $\displaystyle 131 \div 8 \:=\:15,\;r\,11$
I assume you simply mean that x is some integer times 8, plus 11 (which would actually give a remainder of 11- 8= 3).
Okay, let x be that number. Then we require that x- 8i= 11 for some integer, i, that x- 12j= 11 for some integer, j, and that x- 15k= 11 for some integer k. That last equation gives x= 11+ 15k.
Putting x= 11+ 15k into the second equation, 11+ 15k- 12j= 11 so that 12j= 15k or $\displaystyle j= \frac{5}{4}k$. Since j must be an integer, k must be a multiple of 4.
Putting x= 11+ 15k into the first equation, 11+ 15k- 8i= 11 so that 8i= 15k so $\displaystyle i= \frac{15}{8}k$. Since i must be an integer, k must be a multiple of 8. Since that is also a multiple of 4, the smallest value we can have for k is 8.
That gives x= 11+ 15k= 11+ 15(8)= 11+ 120= 131.
Note that, with k= 8, $\displaystyle j= \frac{5}{4}(8)= 10$. That gives x= 11+ 12(10)= 11+ 120= 131 again.
Also, with k= 8, $\displaystyle i= \frac{15}{8}(8)= 15$. That gives x= 11+ 8(15)= 11+ 120= 131.
As a final check, 131 divided by 8 is 16 with a remainder of 3 which could be interpreted as "15 with a remainder of 8+ 3= 11".
131 divided by 12 is 10 with a remainder of 11.
131 divided by 15 is 8 with a remainder of 11.
Too slow again!