Let N = 2^3 \cdot 3^a \cdot 5^b, where a and b are positive integers. If N has exactly 60 … — GMAT

Let $N = 2^3 \cdot 3^a \cdot 5^b$ , where $a$ and $b$ are positive integers. If $N$ has exactly 60 positive integer divisors that are multiples of 10 and exactly 40 positive integer divisors that are multiples of 15, how many positive integer divisors of $N$ are multiples of 12?

A
20
22Answer
C
24
D
33
E
66

Answer

The correct answer is 22. From the problem statement, divisors divisible by 10 require

x \in \{1, 2, 3\}

y \in \{0, \dots, a\}

, and

z \in \{1, \dots, b\}

, leading to

3(a+1)b = 60

, or

(a+1)b = 20

. Divisors divisible by 15 require

x \in \{0, 1, 2, 3\}

y \in \{1, \dots, a\}

, and

z \in \{1, \dots, b\}

, leading to

4ab = 40

, or

ab = 10

. Substituting

ab = 10

into

ab + b = 20

yields

b = 10

and

a = 1

. To be a multiple of

12 = 2^2 \cdot 3^1

, a divisor must have

x \in \{2, 3\}

(2 choices),

y \in \{1\}

(1 choice), and

z \in \{0, 1, \dots, 10\}

(11 choices). The total number of such divisors is

2 \times 1 \times 11 = 22

Step-by-Step Solution

Set up the equation for positive divisors of

N

that are multiples of 10.

Multiples of 10 must be divisible by

2^1 \cdot 5^1

. Thus, for a divisor

2^x \cdot 3^y \cdot 5^z

, we need

1 \le x \le 3

(3 choices),

0 \le y \le a

(

a+1

choices), and

1 \le z \le b

(

b

choices). The total number of such divisors is

3(a + 1)b = 60

, which simplifies to

(a + 1)b = 20

Divisibility by 10 requires at least one factor of 2 and at least one factor of 5.

Set up the equation for positive divisors of

N

that are multiples of 15.

Multiples of 15 must be divisible by

3^1 \cdot 5^1

. Thus, we need

0 \le x \le 3

(4 choices),

1 \le y \le a

(

a

choices), and

1 \le z \le b

(

b

choices). The total number of such divisors is

4ab = 40

, which simplifies to

ab = 10

Divisibility by 15 requires at least one factor of 3 and at least one factor of 5.

Solve the system of equations for

a

and

b

Expanding

(a + 1)b = 20

yields

ab + b = 20

. Substituting

ab = 10

into this equation gives

10 + b = 20 \implies b = 10

. Since

ab = 10

, we find

a(10) = 10 \implies a = 1

Determining the exact exponents

a

and

b

allows us to fully specify

N = 2^3 \cdot 3^1 \cdot 5^{10}

Calculate the number of divisors of

N

that are multiples of 12.

Multiples of 12 must be divisible by

12 = 2^2 \cdot 3^1

. For a divisor

2^x \cdot 3^y \cdot 5^z

, we require

2 \le x \le 3

(2 choices:

2^2, 2^3

1 \le y \le a = 1

(1 choice:

3^1

), and

0 \le z \le b = 10

(

b + 1 = 11

choices:

5^0, 5^1, \dots, 5^{10}

). Total multiples of 12 =

2 \times 1 \times 11 = 22

Applying the fundamental counting principle to the constrained prime factor exponent ranges.

Key Concept

Divisor counting formula under prime factor constraint conditions

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