sorry i intended to apply the spoiler

Are u sure its 3920 in the denominator?

yessir

yes

i am with ramya on this !

I am getting 13 as the answer ! so either i am wrong in wat i am doing or ! i guess will wait for some more replies

Vishu

haha i was getting that for a couple tries too, thats why i posted this. the source is a blog, so maybe the oa is not correct.

p*p*p / 2^4 * 5 * 7^2 = p*p*p / 4^2 *5 * 7^2 4*5*7=140, digits together is 5.

thoughts on this?

hey

I think this works ! but i am just wandering ! wat should be the ideal approach to this kinda problem in real exam scenario ! it was more or less a trial method!

We just cant do this ! as of now we know the OA !! sooooo if anyone could tell wat is the first thing to approach this problem


vishu

i believe this is a classic example of breaking down a number into its primes.

if anyone believes the oa is wrong please advise.

what i dont understand, is why the number is broken down into 4^2 5 and 7, rather than 2^4 5 and 7, as that would give us a different answer.

mberkowitz wrote: if anyone believes the oa is wrong please advise.

what i dont understand, is why the number is broken down into 4^2 5 and 7, rather than 2^4 5 and 7, as that would give us a different answer.

I think OA is perfect.

3920 = 2^4 * 5 * 7^2
P^3 could be 2^6 * 5^3 * 7^3

P = 2*2*5*7 = 140

sum of digits = 1+4 = 5

If you take

P^3 = 4^3 * 5 * 7

P = 4*5*7 = 140

sum of digits = 1+4 = 5


Hope this helps.

Would someone mind explaining this in a little more detail. I got to -

3920 = 2^4*5*7^2

How do you get from here to 4*5*7 ??

Thanks

parallel_chase wrote:

mberkowitz wrote: if anyone believes the oa is wrong please advise.

what i dont understand, is why the number is broken down into 4^2 5 and 7, rather than 2^4 5 and 7, as that would give us a different answer.

I think OA is perfect.

3920 = 2^4 * 5 * 7^2
P^3 could be 2^6 * 5^3 * 7^3

P = 2*2*5*7 = 140

sum of digits = 1+4 = 5

If you take

P^3 = 4^3 * 5 * 7

P = 4*5*7 = 140

sum of digits = 1+4 = 5


Hope this helps.

Hi parallel_chase,
I dint quite understand how you got

P^3 could be 2^6 * 5^3 * 7^3

from the prime factorization of 3920

3920 = 2^4 * 5 * 7^2

Is the logic like this:
Raise the powers of each of the prime factors such that each power is a nearest multiple of 3. If I use that logic, then I get your answer.

I am not sure if that is indeed the logic??

thanks
-V

This was very confusing.
is this a real GMAT question, I wonder?

parallel_chase wrote:

mberkowitz wrote: if anyone believes the oa is wrong please advise.

what i dont understand, is why the number is broken down into 4^2 5 and 7, rather than 2^4 5 and 7, as that would give us a different answer.

I think OA is perfect.

3920 = 2^4 * 5 * 7^2
P^3 could be 2^6 * 5^3 * 7^3

P = 2*2*5*7 = 140

sum of digits = 1+4 = 5

If you take

P^3 = 4^3 * 5 * 7

P = 4*5*7 = 140

sum of digits = 1+4 = 5


Hope this helps.

Absolutely agree with the reasoning but question asks about the sum of the digits, not about the sum of the prime factors.
minimum p is 2^6 * 5^3 * 7^3 = 2744000
the sum of the integers is 25
it also can be 5488009, the sum of the integers is 34
This shows that either OA is incorrect or question is incorrect

i believe you have confused minimum p with minimum p^3