If a and b are positive integers, what is the remainder when ab is

Question

If a and b are positive integers, what is the remainder when ab is divided by 40?
 
(1) b is 60% greater than a.
 
(2) Each of  a^2b  and  ab^2  is divisible by 40.

Bunuel · Accepted Answer

If a and b are positive integers, what is the remainder when ab is divided by 40? 
 
(1) b is 60% greater than a -->  b=1.6a -->  b=\frac{8}{5}a  -->  \frac{b}{a}=\frac{8}{5}  --> b is a multiple of 8 (8x) and a is a multiple of 5 (5x) --> ab=5x*8x=40x^2. The remainder when  ab=40x^2  is divided by 40 is 0. Sufficient.
 
(2) Each of a^2 *b and a*b^2 is divisible by 40. If a=2 and b=10, then ab=20 and the the remainder is also 20 but if a=b=40, then ab=40^2 and the remainder si 0. Not sufficient.

Answer: A.

kingflo · Answer

Could we also pick numbers for (1)?

(1) b = 1,6a 
--> pick smart numbers: b=10 , a=16 --> 10*16 / 40 = 4 + R0 --> Sufficient

Right?

Bunuel · Answer

This won't work in all cases. If the question were "what is the remainder when ab is divided by 80?", then with your numbers you'd still get that the remainder is 0, and the statement is sufficient. But this would be wrong: if b=5 and a=5, then ab=40 and the remainder is 40 not 0.

Hope it's clear.

Jerry1982 · Answer

Hi Bunuel, 
i got this question wrong,but can you poiunt out th eflaw in my reasoning here, 
what i did was i calculated that b= 8* a /5 and then ab = (8 * a^2 )/5. now if i divide this new ab by 40 i am left with a^2/25. after this i plugged numbers for a and got different remainders i.e if a^2 = 1, 4, 9, 16 etc , i will have different remainders therefore S1 is insufficient.

Bunuel · Answer

From b= 8* a /5 (b/a = 8/5) it follows that a must be a multiple of 5, so you should plug only multiples of 5 there.

vivekgautam1 · Answer

Isn't it possible to actually pick numbers here for (1) ?
We know both 'a' and 'b' have to be positive integers, and b is always 1.6 times of a.
So, all integer combinations which satisfy the condition - b=1.6a can help us.

a --> b = 1.6 a
5 --> 8
10 --> 16
15 --> 24 
and likewise, and in all cases remainder will be 0. Right ?

energetics · Answer

You can pick numbers as long as you make sure to test a few combinations, what Bunuel was saying was that if the divisor was 80 we'd get R40 using the numbers 8*5 while we would get R0 using the next set of multiples, 16*10. So you can't just pick one set and assume it will be sufficient in remainder questions since there's often some pattern with different remainders, just not here.

sm1510 · Answer

Can someone pls tell me where am I going wrong with the second statement?
2) Each of  a^2b  and  ab^2  is divisible by 40.

That is: 
 a^2 b = 40x   and 
 ab^2 = 40y  
Subtracting the two equations
 a^2 b - ab^2 = 40x - 40y  
 ab(a - b) = 40 ( x - y)  
 ab = \frac{ }{(a-b)}

This tells us ab is a multiple of 40
So if ab is divided by 40, the remainder would be 0

Insights would be appreciated.

Thanks!

labbasl · Answer

The equation is kind of flawed here, what happens if a=b? you are essentially dividing by 0 which is undefined.

sm1510 · Answer

Do you mean we are not allowed to subtract the two equations ?

Posted from my mobile device

pulkit sareen · Answer

Is there any other way of proving the insufficiency of Statement 2 ? Also how do we specifically choose these numbers? I did not get these numbers in my mind when I was trying this out.

soudipsengupta · Answer

ab = \frac{ }{(a-b)}

This does not necessarily mean ab is a multiple of 40. Observe you are dividing the Numerator (40K) by a-b. a-b could be a factor of 40, thus reducing the 40 in the numerator:

e.g. a=10, b=2; a-b=8; 
therefore ab=5(x-y) -> Not always a multiple of 40

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If a and b are positive integers, what is the remainder when ab is

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