Let K = 2^a \cdot 3^4 \cdot 7^b, where a and b are positive integers. If K has exactly 48 … — GMAT

Let $K = 2^a \cdot 3^4 \cdot 7^b$ , where $a$ and $b$ are positive integers. If $K$ has exactly $48$ positive integer divisors that are multiples of $6$ , and exactly $30$ positive integer divisors that are multiples of $14$ , what is the total number of positive integer divisors of $K$ ?

A
42
B
56
C
60
70Answer
E
84

Answer

The total number of positive integer divisors of

K

is 70.

The prime factorization of

K

2^a \cdot 3^4 \cdot 7^b

. Any positive divisor of

K

takes the form

2^x \cdot 3^y \cdot 7^z

, where

0 \le x \le a

0 \le y \le 4

, and

0 \le z \le b

. A divisor is a multiple of

6 = 2^1 \cdot 3^1

x \ge 1

y \ge 1

, and

z \ge 0

. The number of such divisors is

a \times 4 \times (b+1) = 48

, which simplifies to

a(b+1) = 12

. A divisor is a multiple of

14 = 2^1 \cdot 7^1

x \ge 1

y \ge 0

, and

z \ge 1

. The number of such divisors is

a \times 5 \times b = 30

, which simplifies to

ab = 6

. Substituting

ab = 6

into

ab + a = 12

gives

6 + a = 12 \implies a = 6

, which means

b = 1

. The total number of positive integer divisors of

K = 2^6 \cdot 3^4 \cdot 7^1

(6+1)(4+1)(1+1) = 7 \times 5 \times 2 = 70

Step-by-Step Solution

Express the condition for divisors of

K

being multiples of 6 in terms of exponents.

a \cdot 4 \cdot (b+1) = 48 \implies a(b+1) = 12

A divisor of

K = 2^a \cdot 3^4 \cdot 7^b

has the form

2^x \cdot 3^y \cdot 7^z

with

0 \le x \le a

0 \le y \le 4

0 \le z \le b

. For it to be a multiple of

6 = 2^1 \cdot 3^1

, we must have

x \ge 1

(

a

choices),

y \ge 1

(

4

choices), and

z \ge 0

(

b+1

choices).

Express the condition for divisors of

K

being multiples of 14 in terms of exponents.

a \cdot 5 \cdot b = 30 \implies ab = 6

For a divisor to be a multiple of

14 = 2^1 \cdot 7^1

, we must have

x \ge 1

(

a

choices),

y \ge 0

(

5

choices), and

z \ge 1

(

b

choices).

Solve the system of equations for the positive integer exponents

a

and

b

a = 6

and

b = 1

Expanding

a(b+1) = 12

gives

ab + a = 12

. Substituting

ab = 6

yields

6 + a = 12 \implies a = 6

. Then

6b = 6 \implies b = 1

Calculate the total number of positive integer divisors of

K = 2^6 \cdot 3^4 \cdot 7^1

(6+1)(4+1)(1+1) = 7 \times 5 \times 2 = 70

The total number of divisors of a prime factorization

p_1^{e_1} p_2^{e_2} \dots p_k^{e_k}

is given by

(e_1 + 1)(e_2 + 1) \dots (e_k + 1)

Key Concept

Counting Divisors using Prime Factor Exponents

Estimated Time:2m 0s

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Step-by-Step Solution

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