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# What is the sum of ALL the factors of 49?

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Math Expert
Joined: 02 Sep 2009
Posts: 53066
What is the sum of ALL the factors of 49?  [#permalink]

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16 Nov 2016, 03:53
00:00

Difficulty:

5% (low)

Question Stats:

82% (00:43) correct 18% (00:38) wrong based on 199 sessions

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What is the sum of ALL the factors of 49?

A. 7
B. 8
C. 50
D. 51
E. 57

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Re: What is the sum of ALL the factors of 49?  [#permalink]

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16 Nov 2016, 05:06
2
2
When an integer n can be expressed as $$a^p*b^q*c^r$$, where a,b,c are prime factors of n and p,q,r are their powers,
the sum of the factors of n can be expressed as $$\frac{(a^p+1-1)*(b^q+1-1)*(c^r+1-1)}{(a-1)*(b-1)*(c-1)}$$

49=$$7^2$$
Applying the formula, $$\frac{(7^2+1-1)}{(7-1)=(7^3-1)/6}$$ = 57
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Senior Manager
Joined: 06 Jun 2016
Posts: 258
Location: India
Concentration: Operations, Strategy
Schools: ISB '18 (D)
GMAT 1: 600 Q49 V23
GMAT 2: 680 Q49 V34
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Re: What is the sum of ALL the factors of 49?  [#permalink]

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16 Nov 2016, 05:42
2
Bunuel wrote:
What is the sum of ALL the factors of 49?

A. 7
B. 8
C. 50
D. 51
E. 57

Factors of 49 = 1, 7, 49
Sum= 57
E
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Re: What is the sum of ALL the factors of 49?  [#permalink]

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16 Nov 2016, 07:31
1
Bunuel wrote:
What is the sum of ALL the factors of 49?

A. 7
B. 8
C. 50
D. 51
E. 57

Factors of 49 are 1 , 7 & 49

Sum of all the factors is 1 + 7 + 49 = 57

Hence, answer will be (E) 57

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Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
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Re: What is the sum of ALL the factors of 49?  [#permalink]

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16 Nov 2016, 07:42
1
Bunuel wrote:
What is the sum of ALL the factors of 49?

A. 7
B. 8
C. 50
D. 51
E. 57

what are the factors of 49?
1, 7, 49

1+7+49 = 57
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Joined: 14 Oct 2013
Posts: 14
Re: What is the sum of ALL the factors of 49?  [#permalink]

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16 Nov 2016, 07:49
Factors of 49 --> 1, 7 and 49

Sum of factors = 1 + 7 + 49 = 57

Option E
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Re: What is the sum of ALL the factors of 49?  [#permalink]

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27 Aug 2017, 04:07
Applying the formula here (7^(2+1) -1)/(7-1) = (7^3 -1)/6

= 342/6 =57
Manager
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What is the sum of ALL the factors of 49?  [#permalink]

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31 Aug 2017, 04:40
1
Steps:

1. Find the prime factors to the number in question

2. Sub in the formula below

$$\frac{[a^{(n+1)} -1][b^{(n+1)} -1][c^{(n+1)} -1]...}{[a-1][b-1][c-1]...}$$

So, referring back to the question:

1. Find the prime factors of 49, which is $$7^2$$

2. Subbing in the formula, we get:

=$$\frac{[7^{(2+1)} -1]}{[7-1]}$$

= $$\frac{[343-1]}{6}$$ (we get 343 because $$7^3$$= 343)

= 342/6

= 57

Now, if there are more prime factors to a given number, we could substitute each prime factor in place of a or b or c and so on in the formula above.

Hope that helps!
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Re: What is the sum of ALL the factors of 49?  [#permalink]

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13 Oct 2018, 00:41
factors of 49 are=1,7,49
$$Sum=1+7+49=57$$
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Re: What is the sum of ALL the factors of 49?   [#permalink] 13 Oct 2018, 00:41
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