32^232 + 17^232

Question

32^232 + 17^232 is definetly divisible by:

a. 49 
 b. 15 
c. 49 & 15
d. none of these

subhashghosh · Answer

The answer is D, none of these.

This is because 32^232 is even with only even factors, while 49 and 15 have only odd factors.

Also, This is because 17^232 is co-prime with 49 and 15 (although this step is unnecessary).

toughmat · Answer

Can u pls explain what is Coprime.
Also 32^232 is even and 17^232 is odd evn+odd is odd

subhashghosh · Answer

Co-Prime means numbers that do not have a common factor apart from 1, like 2 and 3, 11 and 12 etc.

I did not get what you mean by this :

Also 32^232 is even and 17^232 is odd evn+odd is odd

sudhir18n · Answer

has to be D
as subhash said.

other approach is CRT

32^232+ 17^232 mod 15 = 16^58+ 16^58 mod 15 = 1^58 + 1^58 mod 15 = 2 mod 15 = 2
similar approach for 49.

other good to know rules
a^n - b^n is always divisible by a-b 
a^n - b^n is always divisible by a+b if n is even
a^n + b^n is always divisible by a+b if n is odd

jamifahad · Answer

Last digit of 32^232 is same as last digit of 2^4 which is 6.
Last digit of 17^232 is same as last digit of 7^4 which is 1.
Adding last digit of 32^232 + 17 ^232 = 6+1=7
Now the resulting number will be of the form xyz7 which can never be divisible by 15. SO b and c is out.

Now i am not very sure how to eliminate 49. Since 49*3= 147, 32^232 + 17^232 could end in 7 and could be divisible by 7.

IanStewart · Answer

This question is clearly not a GMAT question. While you can use units digits to show that 32^232 + 17^232 is not divisible by 5 (and thus by 15), to determine that 32^232 + 17^232 is not divisible by 49 you'd need an advanced understanding of modular arithmetic, and that just isn't tested on the GMAT (you can use the fact that 17^2 and 32^2 give the same remainder when divided by 49 to see that 32^232 + 17^232 won't be divisible by 49). I'm sure this is an IIM question, particularly because it has only four answer choices.

I think there are some misunderstandings in the above. Let's take a much simpler example (don't use this question as a practice question - it's not a good GMAT question). If I ask

2^6 + 3^6 is divisible by:
I) 3
II) 13
III) 61

you'll notice that 2^6 has only even factors, 3^6 has only odd factors, and 3^6, 13 and 61 are relatively prime (share no prime divisors). In these respects, this question is just like the one in the original post. Those facts are all irrelevant; they don't help us to answer the question. 2^6 + 3^6 *is* divisible by 13 and by 61 - you can confirm with a calculator.

In the above question the one answer you *could* rule out immediately is 3. When we add 2^6 and 3^6, we are adding a multiple of 3 to a number which is *not* a multiple of 3. When we do this, we can *never* get a multiple of 3. So it is in that case that common divisors can be relevant. For example 32^232 + 17^232 could not possibly be divisible by 17, since we are adding a multiple of 17 to a non-multiple of 17. But you can't rule out any other odd primes on that basis. I don't know what divisors 32^232 + 17^232 has, but they are absolutely certain to be relatively prime with 32 and 17.

Units digits won't tell you anything about whether a number is divisible by 7. Multiples of 7 can end in anything at all (just list the first ten positive multiples of 7 and you will see you get all ten possible units digits: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70). Units digits will tell you when you can divide by 2, 5 or 10, but otherwise don't tell you much about divisibility.

32^232 + 17^232

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