Question: Find the sum of the sum of even divisors of 96 and the sum of odd divisors of 3600?

A) 639
B) 739
C) 735
D) 651
E) 589

Correct Answer: D
Solution and Explanation:
Approach Solution 1:

The problem statement asks to find out the sum of the sum of even divisors of 96 and the sum of odd divisors of 3600.
If the number is \(p^a*t^b\), the sum is \((p^0+p^1+....p^a)(t^0+t^1+....+t^b.…\)

Here the two numbers are

1) 96 which can be derived as \(2^5*3^1\)

Sum of even divisors = total - sum of odd factors.
Total = \((2^0+2^1+2^2+2^3+2^4+2^5)(3^0+3^1)\) = 63 ∗ 4 = 252
Sum of odd factors= \(3^0+3^1\) = 4
Sum of even divisors= 252 - 4 = 248

2) odd divisors of 3600
3600= \(2^4*3^2*5^2\)

Odd divisors sum = \((3^0+3^1+3^2)(5^0+5^1+5^2) \)= 13 ∗ 31 = 403
Therefore, the sum of the sum of even divisors of 96 and the sum of odd divisors of 3600
= 248 + 403 = 651

Approach Solution 2:

The problem statement asks to find out the sum of the sum of even divisors of 96 and the sum of odd divisors of 3600.

96 = \(2^5 * 3^1\)

Sum of EVEN divisor of 96 = \((2^1 + 2^2 + 2^3 + 2^4 +2^5) * (3^0 +3^1)\)

(we are not adding \(2^0\) term as when we multiply with any odd term we get odd term only )
\(=2(2^5-1)/(2-1) * (3^2-1)/(3-1) = 62 * 4 = 248\)
\(3600 = 2^4 * 3^2 * 5^2\)

Sum of the ODD divisor of 96 = \((3^0 +3^1 +3^2) * (5^0+5^1+5^2)\)

(here for odd divisor we consider only odd term since we know even * odd =even)
\(= (3^3-1)/(3-1) * (5^3-1)/(5-1) =13 * 31 = 403\)

Therefore, SUM is 248 + 403 = 651

Hence, the sum of the sum of even divisors of 96 and the sum of odd divisors of 3600 = 651

Approach Solution 3:

The problem statement asks to find out the sum of the sum of even divisors of 96 and the sum of odd divisors of 3600.
We can calculate the sum of divisors with the use of this formula:
If n = \(a^x*b^y\), then the sum of divisors of n = \(a^{(x+1)}-1/(a-1) * b^{(y+1)}-1/(b-1)\)

Using this we can calculate for 96.
96 = \(2^5*3^1\)

Sum (D) = \((2^6-1)/(2-1) * (3^2-1)/(3-1)\) = 63 ∗ 4 = 252
Now to calculate the sum of (even D), we need to subtract the odd ones. Only 1 and 3 are the odd ones.
Sum (even D) of 96 = 252 - 4 = 248

If n = \(a^x*b^y*c^z\),
then the sum of divisors of n = \(a^{(x+1)}-1/(a-1) * b^{(y+1)}-1/(b-1)*c^{(z+1)}-1/(c-1)\)

Using this we can calculate for 96
3600 = \(2^4 * 3^2 * 5^2\)

Sum (odd D) = \((3^3-1)/(3-1)*(5^3-1)/(5-1)\) = 13 ∗ 31 = 403

Therefore, SUM is 248 + 403 = 651
Hence, the sum of the sum of even divisors of 96 and the sum of odd divisors of 3600 = 651

“Find the sum of the sum of even divisors of 96 and the sum of odd divisors”- is a topic of the GMAT Quantitative reasoning section of the GMAT exam. This topic has been taken from the book “Foundations of GMAT Math”. The GMAT Problem Solving questions enable the candidates to analyse the data in order to interpret numerical problems. GMAT Quant practice papers help the candidates to go through varieties of questions that will enable them to improve their mathematical understanding.

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