Let N = 2^a \cdot 3^b \cdot 5^c, where a, b, and c are positive integers such that N is a … — GRE General Test

Let $N = 2^a \cdot 3^b \cdot 5^c$ , where $a$ , $b$ , and $c$ are positive integers such that $N$ is a multiple of $8$ . If $N^2$ has $105$ distinct positive divisors and $\frac{N}{6}$ has $12$ distinct positive divisors, how many distinct positive divisors does $10N$ have?

A
30
B
36
45Answer
D
48
E
60

Answer

The number of distinct positive divisors of

10N

is 45.

The correct answer is 45. Given

N = 2^a \cdot 3^b \cdot 5^c

, the number of positive divisors of

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

(2a+1)(2b+1)(2c+1) = 105

. The unique factor decomposition of 105 into three factors greater than 1 is

3 \times 5 \times 7

. Because

N

is a multiple of 8,

a \ge 3

, so

2a+1 \ge 7

, forcing

2a+1 = 7

, which means

a = 3

. The remaining factor set

\{2b+1, 2c+1\} = \{3, 5\}

gives two cases for

(b, c)

(2, 1)

(1, 2)

. For

\frac{N}{6} = 2^{a-1} \cdot 3^{b-1} \cdot 5^c

, the divisor count is

a \cdot b \cdot (c+1) = 3 \cdot b \cdot (c+1) = 12

, so

b(c+1) = 4

. Substituting

(b, c) = (2, 1)

yields

2 \times 2 = 4

, which satisfies this relation. Therefore,

(a, b, c) = (3, 2, 1)

. Now

10N = 2^{a+1} \cdot 3^b \cdot 5^{c+1} = 2^4 \cdot 3^2 \cdot 5^2

. Its total number of positive divisors is

(4+1)(2+1)(2+1) = 5 \times 3 \times 3 = 45

Step-by-Step Solution

Express the divisor count formula for

N^2

d(N^2) = (2a+1)(2b+1)(2c+1) = 105

For an integer with prime factorization

p_1^{e_1} p_2^{e_2} \cdots

, the number of divisors is

(e_1+1)(e_2+1)\cdots

. Here

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

Determine the value of

a

using the divisibility condition

a = 3

Since

N

is a multiple of

8 = 2^3

, we must have

a \ge 3

, which implies

2a+1 \ge 7

. The prime factorization of

105

into three odd factors greater than

1

3 \times 5 \times 7

. Thus,

2a+1 = 7

, giving

a = 3

Solve for exponents

b

and

c

using the divisor count of

\frac{N}{6}

b = 2

and

c = 1

Since

\frac{N}{6} = 2^{a-1} \cdot 3^{b-1} \cdot 5^c

, its divisor count is

a \cdot b \cdot (c+1) = 3b(c+1) = 12

, so

b(c+1) = 4

. From step 2,

\{2b+1, 2c+1\} = \{3, 5\}

, so either

(b, c) = (2, 1)

(b, c) = (1, 2)

. Testing

(b, c) = (2, 1)

gives

2(1+1) = 4

, which satisfies the equation.

Calculate the prime factorization and divisor count of

10N

d(10N) = (4+1)(2+1)(2+1) = 45

Since

10N = 2 \cdot 5 \cdot (2^3 \cdot 3^2 \cdot 5^1) = 2^4 \cdot 3^2 \cdot 5^2

, applying the divisor formula yields

5 \times 3 \times 3 = 45

Key Concept

Divisor Count Formula & Prime Factorization Constraints

Estimated Time:2m 30s

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