Number of factors

To count the number of positive factors of an integer:

1) Prime-factorize the integer
2) Add 1 to each exponent
3) Multiply

For example:
72 = 2Â³ * 3Â².
Adding 1 to each exponent and multiplying, we get (3+1)*(2+1) = 12 positive factors.

Here's why:
To determine how many factors can be created from 72 = 2Â³ * 3Â², we need to determine the number of choices we have of each prime factor and to count the number of ways these choices can be combined:

For 2, we can use 2â�°, 2Â¹, 2Â², or 2Â³, giving us 4 choices.
For 3, we can use 3â�°, 3Â¹, or 3Â², giving us 3 choices.

Multiplying the number of choices we have of each factor, we get 4*3 = 12 positive factors.

What is the number of factors of 4N (where N is a natural number)?

!)2N is having 20 factors.

2)3N is having 15 factors

Statement 1:
Case 1: 2N = 2xâ�¹, where N = xâ�¹ and xâ‰ 2
Since 2N = 2Â¹xâ�¹, the number of factors for 2N = (1+1)(9+1) = 20.

Case 2: 2N = 2xâ�´y, where N = xâ�´y, xâ‰ 2, yâ‰ 2 and xâ‰ y.
Since 2N = 2Â¹xâ�´yÂ¹, the number of factors for 2N = (1+1)(4+1)(1+1) = 20.

Since each option for N will yield a different number of factors for 4N, INSUFFICIENT.

Other cases:

Case 3: 2N = 2(2Â²xâ�´) = 2Â³xâ�´, where N = 2Â²xâ�´ and xâ‰ 2
Since 2N = 2Â³xâ�´, the number of factors for 2N = (3+1)(4+1) = 20.

Case 4: 2N = 2(2Â³xÂ³) = 2â�´xÂ³, where N = 2Â³xÂ³ and xâ‰ 2
Since 2N = 2â�´xÂ³, the number of factors for 2N = (4+1)(3+1) = 20.

Case 5: 2N = 2(2Â³xy) = 2â�´xy, where N = 2Â³xy and xâ‰ 2, xâ‰ 2, yâ‰ 2 and xâ‰ y
Since 2N = 2â�´xÂ¹yÂ¹, the number of factors for 2N = (4+1)(1+1)(1+1) = 20.

Case 6: 2N = 2(2â�¸x) = 2â�¹x, where N = 2â�¸x and xâ‰ 2
Since 2N = 2â�¹xÂ¹, the number of factors for 2N = (9+1)(1+1) = 20.

Case 7: 2N = 2(2Â¹â�¸) = 2Â¹â�¹, where N = 2Â¹â�¸
Since 2N = 2Â¹â�¹, the number of factors for 2N = 19+1 = 20.

Statement 2:
Only three cases are possible:
Case 1: 3N = 3(3xâ�´) = 3Â²xâ�´, where N = 3xâ�´ and xâ‰ 3
Since 3N = 3Â²xâ�´, the number of factors for 3N = (2+1)(4+1) = 15.

Case 2: 3N = 3(3Â³xÂ²) = 3â�´xÂ², where N = 3Â³xÂ² and xâ‰ 3
Since 3N = 3â�´xÂ², the number of factors for 3N = (4+1)(2+1) = 15.

Case 3: 3N = 3(3Â¹Â³) = 3Â¹â�´, where N = 3Â¹Â³.
Since 3N = 3Â¹â�´, the number of factors for 3N = 14+1 = 15.

Since each option for N will yield a different number of factors for 4N, INSUFFICIENT.

Statements combined:
Only the value in blue (N = xâ�´y = 3xâ�´) satisfies both statements.
Since N=3xâ�´, 4N = 4(3xâ�´) = 2Â²3Â¹xâ�´, with the result that the number of factors for 4N = (2+1)(1+1)(4+1) = 30.

The correct answer is C.

This problem seems too complex for the GMAT.
What is the source?

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