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Bunuel
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It’s a small enough number that you can list out the factors and then add them up to get the answer.

However in cases of larger numbers with several prime bases, the approach is as follows:

Ex: for the number 36 = (2)^2 * (3)^2, if we were to add up all the factors, we need to first understand the following:

Each unique positive factor can be “made” by using a unique combination of the prime factors.

For 2 we have as available options:
(2)^0 ——— (2)^1 ———(2)^2


For 3 we have as available options:
(3)^0 ———- (3)^1 ——— (3)^2

We can put together each unique factor by setting up all the different arrangements in which we pick 1 of the 3 available options for each prime base. Then we can SUM these factors:

(2)^0 * (3)^0 + (2)^0 *(3)^1 + (2)^0 *(3)^2

+ (2)^1 * (3)^0 + (2)^1 * (3)^1 + (2)^1*(3)^2

+ (2)^2 * (3)^0 + (2)^2 * (3)^1 + (2)^2 * (3)^2


Then for each line, we can take out the common factor or (2) to the given power:


(2)^0 * [ (3)^0 + (3)^1 + (3)^2]
+
(2)^1 * [ (3)^0 + (3)^1 + (3)^2]
+
(2)^2 * [ (3)^0 + (3)^1 + (3)^2]

Then you can take as common the expression [ (3)^0 + (3)^1 + (3)^2 ]


[ (3)^0 + (3)^1 + (3)^2] * [ (2)^0 + (2)^1 (2)^2 ]

[ 1 + 3 + 9] * [1 + 2 + 4] = [13] * [7] = 91

And

1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 + 36 = 91

So to find the sum of all the unique positive factors of a given number:

(1st)Prime Factorize the number

(2nd) Add up all the Prime Base terms for each Prime Factor starting with the 0th Power (i.e., 1)

(3rd) Find the product of each of those SUMS to get the Sum of all the unique positive factors of the given number (note: can use the Sum of a Finite Geometric Sequence formula for larger amounts of bases)


For 49

49 = (7)^2

[ (7)^0 + (7)^1 + (7)^2 ] = sum of all positive factors

[ 1 + 7 + 49 ] = 57

(E) 57

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A perfect square has three factors
Factors of 49 :1,7,49
Sum 1+7+49;57
Option E

Bunuel
What is the sum of ALL the positive factors of 49?

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

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Factors of 49= 1, 7, 49
therefore their sum is 1+7+49= 57
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