If s is the sum of all integers

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If s is the sum of all integers

by stevecultt » Mon Jun 12, 2017 4:20 am
If s is the sum of all integers from 1 to 30, inclusive, what is the sum of all the factors of s?

(A) 303
(B) 613
(C) 675
(D) 737
(E) 768

OA E

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by DavidG@VeritasPrep » Mon Jun 12, 2017 4:35 am
stevecultt wrote:If s is the sum of all integers from 1 to 30, inclusive, what is the sum of all the factors of s?

(A) 303
(B) 613
(C) 675
(D) 737
(E) 768

OA E
First let's find s. 1 to 30 inclusive is an evenly spaced set, so we know that the average = (high + low)/2 = (30 + 1)/2 = 15.5
Sum = Average * Number of terms = 15.5 * 30 = 465

Prime factorization of 465 = 3 * 5 * 31

Now we can build all of 465's factors from those prime bases: 1, 3, 5, 3*5, 31, 3*31, 5*31, 465
Clean it up: 1, 3, 5, 15, 31, 93, 155, 465

Add 'em to get 768
(or simply see that the units' digit would be '8.'
The answer is E
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by Jay@ManhattanReview » Mon Jun 12, 2017 4:38 am
stevecultt wrote:If s is the sum of all integers from 1 to 30, inclusive, what is the sum of all the factors of s?

(A) 303
(B) 613
(C) 675
(D) 737
(E) 768

OA E
We have 30 consecutive integers starting from 1, thus the median of the numbers would be (1+30)/2 = 31/2. Since consecutive integers from 1 to 30 form an even spaced set, the median of the set = mean of the set.

Thus sum s = Mean x number of integers = (31/2) x 30 = 15x31 = 3x5x31;

Thus, the prime factors of s are: 3, 5 and 31.

Thus, the factors of s are: 1; 3; 5; 31; 3 x 5 ƒ= 15; 3 x 31 ƒ= 93; 5 x 31 ƒ= 155; and 3 x 5 x 31 =ƒ 465.

Thus, the sum of all the factors of s ƒ 1 +‚ 3 +‚ 5 +‚ 31 +‚ 15 ‚+‚ 93 ‚+ 155 ‚+ 465 ƒ= 768.

The correct answer: E

Hope this helps!

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stevecultt wrote:
Mon Jun 12, 2017 4:20 am
If s is the sum of all integers from 1 to 30, inclusive, what is the sum of all the factors of s?

(A) 303
(B) 613
(C) 675
(D) 737
(E) 768

OA E

The sum of all integers from 1 to 30, inclusive, is 30(30 + 1) / 2 = 15(31) = 3 x 5 x 31 = 465.

Recall that the total number of factors of a number is calculated by adding 1 to the exponent of each prime factor of the number and then multiplying those sums together. Since s = 3^1 x 5^1 x 31^1, then 465 has (1 + 1) x (1 + 1) x (1 + 1) = 8 factors. These 8 factors are:

1, 465
3, 155
5, 93
15, 31

Therefore, the sum of all the factors of s is 1 + 3 + 5 + 15 + 31 + 93 + 155 + 465 = 768.

Answer: E

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