Let N = 2^4 \times 3^3 \times 5^2 \times 7. How many positive integer divisors of N are ev… — GMAT

Let $N = 2^4 \times 3^3 \times 5^2 \times 7$ . How many positive integer divisors of $N$ are even, divisible by 15, but not divisible by 9?

Answer: 16

Answer

Any divisor of

N = 2^4 \times 3^3 \times 5^2 \times 7^1

takes the form

2^a \times 3^b \times 5^c \times 7^d

. The condition that the divisor is even requires

a \ge 1

, which gives 4 choices (

a \in \{1, 2, 3, 4\}

). The condition that the divisor is divisible by 15 requires

b \ge 1

and

c \ge 1

. The condition that it is not divisible by 9 requires

b < 2

. Together,

b \ge 1

and

b < 2

mean

b = 1

(1 choice). The condition

c \ge 1

allows

c \in \{1, 2\}

(2 choices). The exponent

d

can be 0 or 1 (2 choices). Multiplying these choices gives

4 \times 1 \times 2 \times 2 = 16

Step-by-Step Solution

Express the general prime factorization of a divisor

d = 2^a \times 3^b \times 5^c \times 7^d

with

0 \le a \le 4

0 \le b \le 3

0 \le c \le 2

0 \le d \le 1

Any divisor of

N

must consist only of the prime factors of

N

up to their respective maximum powers.

Apply the condition for even numbers

a \in \{1, 2, 3, 4\}

(4 choices)

An even integer must contain at least one factor of 2.

Apply the condition for divisibility by 15 and non-divisibility by 9

b = 1

(1 choice) and

c \in \{1, 2\}

(2 choices)

Divisibility by

15 = 3 \times 5

requires

b \ge 1

and

c \ge 1

. Non-divisibility by

9 = 3^2

requires

b < 2

. Thus

b

must be exactly 1.

Apply the condition for the exponent of 7

d \in \{0, 1\}

(2 choices)

There are no restrictions given for prime factor 7.

Calculate the product of choices

4 \times 1 \times 2 \times 2 = 16

By the fundamental counting principle, the total number of valid combinations of prime exponents is the product of the number of choices for each exponent.

Key Concept

Counting Divisors with Prime Factorization Restrictions

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