1. ## Division and remainder

What is the smallest number which when divided by 8,12 or 15 leaves remainder of 11 ?

2. ## Re: Division and remainder

Originally Posted by haftakhan
What is the smallest number which when divided by 8,12 or 15 leaves remainder of 11 ?
a number divided by 8 cannot have a remainder of 11

3. ## Re: Division and remainder

But the question states this. Maybe the question is wrong

4. ## Re: Division and remainder

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.

. . $\text{LCM}(8,12,15) + 11 \:=\:120 + 11 \:=\:\boxed{131}$

Check

. . $131 \div 15 \:=\:8,\;r\,11$

. . $131 \div 12 \:=\:10,\;r\,11$

And I suppose we can write:
. . $131 \div 8 \:=\:15,\;r\,11$

5. ## Re: Division and remainder

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 $j= \frac{5}{4}k$. 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 $i= \frac{15}{8}k$. 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, $j= \frac{5}{4}(8)= 10$. That gives x= 11+ 12(10)= 11+ 120= 131 again.

Also, with k= 8, $i= \frac{15}{8}(8)= 15$. 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!