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:
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Check
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And I suppose we can write:
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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 . 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 . 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, . That gives x= 11+ 12(10)= 11+ 120= 131 again.
Also, with k= 8, . 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!