p^3 / 3920

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p^3 / 3920

by mberkowitz » Tue Oct 28, 2008 8:01 pm
if pis the smallest positive integer such that p^3 / 3920 is also an integer, what is the sum of the digits of p?

a) 5 b)7 c)9 d)11 e)13

oa a[spoiler][/spoiler]
Source: — Problem Solving |

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by mberkowitz » Tue Oct 28, 2008 8:04 pm
sorry i intended to apply the spoiler

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by cramya » Tue Oct 28, 2008 8:10 pm
Are u sure its 3920 in the denominator?

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by mberkowitz » Tue Oct 28, 2008 8:14 pm
yessir

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by mberkowitz » Tue Oct 28, 2008 8:14 pm
yes

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by vishubn » Tue Oct 28, 2008 8:21 pm
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

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by mberkowitz » Tue Oct 28, 2008 8:26 pm
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?

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by vishubn » Tue Oct 28, 2008 8:45 pm
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

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by mberkowitz » Tue Oct 28, 2008 8:51 pm
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.

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by parallel_chase » Wed Oct 29, 2008 1:11 am
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.
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by lachlanc » Wed Oct 29, 2008 6:22 am
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

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by vittalgmat » Wed Oct 29, 2008 8:49 am
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

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by tkhalid » Wed Oct 29, 2008 9:28 am
This was very confusing.
is this a real GMAT question, I wonder?

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by 4meonly » Wed Oct 29, 2008 10:57 am
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

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by mberkowitz » Wed Oct 29, 2008 10:59 am
i believe you have confused minimum p with minimum p^3