A positive integer N has the prime factorization N = 3^a \cdot 5^b, where a and b are posi… — GMAT

A positive integer $N$ has the prime factorization $N = 3^a \cdot 5^b$ , where $a$ and $b$ are positive integers. If $N^2$ has exactly 35 positive integer divisors, what is the number of positive integer divisors of $N^3$ ?

Answer: 70

Answer

Since

N = 3^a \cdot 5^b

N^2 = 3^{2a} \cdot 5^{2b}

. The number of positive divisors of

N^2

(2a+1)(2b+1) = 35

. The unique integer factor pair of 35 greater than 1 is

5 \times 7

, so the exponents

a

and

b

must be 2 and 3 (in some order). For

N^3 = 3^{3a} \cdot 5^{3b}

, the exponents are 6 and 9. Therefore, the number of positive divisors of

N^3

(6+1)(9+1) = 70

Step-by-Step Solution

Set up the formula for the number of positive divisors of

N^2

(2a + 1)(2b + 1) = 35

For an integer with prime factorization

p_1^{e_1} p_2^{e_2}

, the total number of divisors is

(e_1 + 1)(e_2 + 1)

Determine the positive integer values of

a

and

b

One exponent is 2 and the other exponent is 3

35 factors into

5 \times 7

. Solving

2a + 1 = 5

gives

a = 2

, and

2b + 1 = 7

gives

b = 3

Calculate the number of divisors of

N^3

(3(2) + 1)(3(3) + 1) = 7 \times 10 = 70

Exponents of

N^3

are

3a = 6

and

3b = 9

, so total divisors equal

(6 + 1)(9 + 1) = 70

Key Concept

Number of positive integer divisors from prime factorization

Estimated Time:1m 30s

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