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What is the sum of ALL the factors of 49?

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

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New post 16 Nov 2016, 04:53
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Re: What is the sum of ALL the factors of 49? [#permalink]

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New post 16 Nov 2016, 06:06
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
So, answer E.
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Re: What is the sum of ALL the factors of 49? [#permalink]

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New post 16 Nov 2016, 06:42
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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New post 16 Nov 2016, 08:31
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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Re: What is the sum of ALL the factors of 49? [#permalink]

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New post 16 Nov 2016, 08:42
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
answer must be E.

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

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New post 16 Nov 2016, 08:49
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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New post 27 Aug 2017, 05:07
Applying the formula here (7^(2+1) -1)/(7-1) = (7^3 -1)/6

= 342/6 =57

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

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New post 31 Aug 2017, 05:40
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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What is the sum of ALL the factors of 49?   [#permalink] 31 Aug 2017, 05:40
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