If n = 2^4 \times 5^3 \times 7^1, how many positive integer divisors of n are divisible by… — GMAT

If $n = 2^4 \times 5^3 \times 7^1$ , how many positive integer divisors of $n$ are divisible by $10$ but are not multiples of $35$ ?

The value 12 is correct because a divisor

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

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

is divisible by 10 when

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

(4 options) and

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

(3 options). To ensure it is not divisible by 35 (

5 \times 7

c

cannot be 1 when

b \ge 1

. Thus

c

must be 0 (1 option). Multiplying the choices yields

4 \times 3 \times 1 = 12

Express the general form of a divisor of

n

Any divisor

d

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

can be written as

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

, where

0 \le a \le 4

0 \le b \le 3

, and

0 \le c \le 1

Prime factorization determines the structure of all positive divisors.

Apply the condition that

d

must be divisible by

10

Since

10 = 2^1 \times 5^1

, we require

a \ge 1

and

b \ge 1

A number is divisible by 10 if and only if it contains at least one factor of 2 and at least one factor of 5.

Apply the condition that

d

must NOT be a multiple of

35

Since

35 = 5^1 \times 7^1

, a multiple of 35 must have

b \ge 1

and

c \ge 1

. Since

b \ge 1

is already satisfied, to avoid being a multiple of 35, we must set

c = 0

c = 1

, then combined with

b \ge 1

, the divisor would automatically be a multiple of 35.

Calculate the total number of valid choices for

(a, b, c)

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

(4 options),

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

(3 options),

c \in \{0\}

(1 option). Total =

4 \times 3 \times 1 = 12

By the Fundamental Counting Principle, multiplying the independent choices gives the number of valid divisors.

Counting Divisors with Prime Factor Constraints

Estimated Time:1m 30s

Question