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Find the sum of divisors of 544 excluding 1 and 544

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Find the sum of divisors of 544 excluding 1 and 544  [#permalink]

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New post 23 Jan 2017, 20:48
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Find the sum of divisors of 544 excluding 1 and 544.
(A) 1089
(B) 545
(C) 589
(D) 1134
(E)1589

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GMAT 1: 780 Q51 V46
Re: Find the sum of divisors of 544 excluding 1 and 544  [#permalink]

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New post 23 Jan 2017, 21:30
We can prime factor 544 to 2*2*2*2*2*17. And we can find all the factors of 544 by considering every way the prime factors can be combined. So:

2=2
2*2=4
2*2*2=8
2*2*2*2=16
2*2*2*2*2=32
17=17
17*2=34
17*2*2=68
17*2*2*2=136
17*2*2*2*2=272
17*2*2*2*2*2=544 (Disregard, since the question stem tells us to)
(We can also think of the factor of 1 as the case in which we don't choose any of the PFs, but we disregard that here since the question stem tells us to)

So we have:

2+4+8+16+32+17+34+68+136+272 = 589

(There are lots of ways to do the addition. Looking for shortcuts, I first saw that the sum would be odd since there was only one odd number in our list. Sadly, this only eliminated one answer. Then I just added up the units digits to find that our sum would end in 9. Sadly, this just eliminated one more answer. Then I added them up out of order, something like this:

(2+8) + (16+4) + (32+68) + (34+136) + 272 + 17=
10 + 20 + 100 + 170 + 272 + 17 =
300 + 272 + 17 = 589

For this question, I think it might have been faster just to add them up the conventional way, but the odd/even trick and the last digit trick often pay off on such questions. Perhaps a combination of estimation and the last digit trick would have been fastest here.)

For more on the logic behind finding all the factors of a number see this video: https://www.youtube.com/watch?v=njePP0ZK2bY
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Re: Find the sum of divisors of 544 excluding 1 and 544  [#permalink]

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New post 23 Jan 2017, 22:20
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saswata4s wrote:
Find the sum of divisors of 544 excluding 1 and 544.
(A) 1089
(B) 545
(C) 589
(D) 1134
(E)1589



544 = ( 2^5 * 17 )

The sum of divisors of 544 =

(2^0 + 2^1 + 2^2 + 2^3 + 2^4 + + 2^5 ) (17^0 + 17 ^1 ) = (1 + 2 + 4 + 8 + 16 + 32 ) (1 + 17)
= 63 * 18 = 1134

Since we have to exclude 1 & 544 , subtract 545 from 1134 = 589 (Answer )

Similarly , for number of divisor. Considering this expression (2^0 + 2^1 + 2^2 + 2^3 + 2^4 + + 2^5 ) (17^0 + 17 ^1 )

6 * 2 = 12

Then subtract 2 from this as we don't have to consider 1 & 544.
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Re: Find the sum of divisors of 544 excluding 1 and 544  [#permalink]

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New post 04 Feb 2019, 19:20
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Re: Find the sum of divisors of 544 excluding 1 and 544   [#permalink] 04 Feb 2019, 19:20
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