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# Find the sum of the sum of even divisors of 96 and the sum of odd divi

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SVP
Joined: 08 Jul 2010
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Find the sum of the sum of even divisors of 96 and the sum of odd divi [#permalink]

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12 Jun 2017, 06:33
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95% (hard)

Question Stats:

49% (02:26) correct 51% (03:46) wrong based on 72 sessions

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Find the sum of the sum of even divisors of 96 and the sum of odd divisors of 3600?

A) 639
B) 739
C) 735
D) 651
E) 589
[Reveal] Spoiler: OA

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Kudos [?]: 2401 [2], given: 51

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Re: Find the sum of the sum of even divisors of 96 and the sum of odd divi [#permalink]

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12 Jun 2017, 07:56
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GMATinsight wrote:
Find the sum of the sum of even divisors of 96 and the sum of odd divisors of 3600?

A) 639
B) 739
C) 735
D) 651
E) 589

Firstly, yet to see a Q testing sum of factors!!
Also the choices don't seem to be as per GMAT, where they are always in ascending or descending order

But if you want to know the method..
if the number is $$p^a*t^b$$ sum is $$p^0+p^1+....p^a)(t^0+t^1+......+t^b$$...
Here the two numbers are
1) 96=2^5*3...
Sum of even divisors= total - sum of odd factors..
Total =$$(2^0+2^1+2^2+2^3+2^4+2^5)(3^0+3^1)=63*4=252$$
Sum of odd factors=$$3^0+3^1=4$$..
Sum of even divisors=252-4=248
2) odd divisors of 3600..
3600=$$2^4*3^2*5^2$$..
Odd divisors sum = $$(3^0+3^1+3^2)(5^0+5^1+5^2)=13*31=403$$..

Sum = 248+403=651..
D
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Re: Find the sum of the sum of even divisors of 96 and the sum of odd divi [#permalink]

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23 Sep 2017, 10:54
chetan2u wrote:
GMATinsight wrote:
Find the sum of the sum of even divisors of 96 and the sum of odd divisors of 3600?

A) 639
B) 739
C) 735
D) 651
E) 589

Firstly, yet to see a Q testing sum of factors!!
Also the choices don't seem to be as per GMAT, where they are always in ascending or descending order

But if you want to know the method..
if the number is $$p^a*t^b$$ sum is $$p^0+p^1+....p^a)(t^0+t^1+......+t^b$$...
Here the two numbers are
1) 96=2^5*3...
Sum of even divisors= total - sum of odd factors..
Total =$$(2^0+2^1+2^2+2^3+2^4+2^5)(3^0+3^1)=63*4=252$$
Sum of odd factors=$$3^0+3^1=4$$..
Sum of even divisors=252-4=248
2) odd divisors of 3600..
3600=$$2^4*3^2*5^2$$..
Odd divisors sum = $$(3^0+3^1+3^2)(5^0+5^1+5^2)=13*31=403$$..

Sum = 248+403=651..
D

Hello Chetan,

Small doubt..

while calculating for 96.. you took even divisors = total - odd
But while calculating for 3600... you directly took odd divisors.

Is this approach correct..?
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Re: Find the sum of the sum of even divisors of 96 and the sum of odd divi [#permalink]

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23 Sep 2017, 12:18
3
KUDOS
Bharath99 wrote:
chetan2u wrote:
GMATinsight wrote:
Find the sum of the sum of even divisors of 96 and the sum of odd divisors of 3600?

A) 639
B) 739
C) 735
D) 651
E) 589

Firstly, yet to see a Q testing sum of factors!!
Also the choices don't seem to be as per GMAT, where they are always in ascending or descending order

But if you want to know the method..
if the number is $$p^a*t^b$$ sum is $$p^0+p^1+....p^a)(t^0+t^1+......+t^b$$...
Here the two numbers are
1) 96=2^5*3...
Sum of even divisors= total - sum of odd factors..
Total =$$(2^0+2^1+2^2+2^3+2^4+2^5)(3^0+3^1)=63*4=252$$
Sum of odd factors=$$3^0+3^1=4$$..
Sum of even divisors=252-4=248
2) odd divisors of 3600..
3600=$$2^4*3^2*5^2$$..
Odd divisors sum = $$(3^0+3^1+3^2)(5^0+5^1+5^2)=13*31=403$$..

Sum = 248+403=651..
D

Hello Chetan,

Small doubt..

while calculating for 96.. you took even divisors = total - odd
But while calculating for 3600... you directly took odd divisors.

Is this approach correct..?

Hi,

its because, even * odd = even.. so, to get all even divisors we subtract the odd divisors , so that the list is complete

suppose, if we take just the even divisors directly, then we will not be considering divisors like 6(2*3), 12(4*3) ... etc...

but for calculating odd divisors, we don't need to do the above because, only odd*odd = odd..

hope you understood it..

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Re: Find the sum of the sum of even divisors of 96 and the sum of odd divi [#permalink]

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23 Sep 2017, 23:36
People have already explained how to solve this problem. Just to add, you can also calculate sum of divisors with this formula.

If n= $$a^x * b^y$$ ,

then sum of divisors of n = $$\frac{a^(x+1)-1}{a-1} * \frac{b^(y+1)-1}{b-1}$$

Using this you can calculate for 96.
96 = $$2^5 * 3^1$$
Sum (D) = $$\frac{(2^6-1)}{2-1} * \frac{(3^2-1)}{3-1}$$

= $$63 * 4$$

= 252
Now to calculate sum of (even D), just subtract the odd ones. Only 1 and 3 are the odd ones.
Sum (even D) of 96 = 252 -4
= 248
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Re: Find the sum of the sum of even divisors of 96 and the sum of odd divi [#permalink]

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25 Sep 2017, 03:41
Bunuel

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Re: Find the sum of the sum of even divisors of 96 and the sum of odd divi [#permalink]

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25 Sep 2017, 03:55
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Re: Find the sum of the sum of even divisors of 96 and the sum of odd divi   [#permalink] 25 Sep 2017, 03:55
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