A positive integer n is a multiple of 6, but is neither a multiple of 4 nor a multiple of … — GMAT

A positive integer $n$ is a multiple of $6$ , but is neither a multiple of $4$ nor a multiple of $9$ . If $n$ has exactly $12$ positive integer divisors and $n^2$ has exactly $45$ positive integer divisors, how many positive integer divisors of $n^2$ are divisible by $n$ ?

Cevap: 12

Cevap

The number of positive integer divisors of

n^2

that are divisible by

n

12

By analyzing the prime factorization

n = 2^1 \cdot 3^1 \cdot p^2

, we find

n^2 = 2^2 \cdot 3^2 \cdot p^4

. For a divisor

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

n^2

to be divisible by

n

, the exponents must satisfy

x \in \{1, 2\}

y \in \{1, 2\}

, and

z \in \{2, 3, 4\}

, yielding

2 \times 2 \times 3 = 12

valid divisors.

Adım Adım Çözüm

Analyze the prime factorization of

n

for prime factors

2

and

3

The exponents of

2

and

3

n

are both

1

Since

n

is a multiple of

6

, it must contain at least one factor of

2

and one factor of

3

. Because

n

is not a multiple of

4

, the exponent of

2

cannot exceed

1

. Because

n

is not a multiple of

9

, the exponent of

3

cannot exceed

1

Determine the remaining prime factors of

n

using the total divisor count of

n

n = 2^1 \cdot 3^1 \cdot p^2

for some prime

p > 3

The number of positive divisors is given by

d(n) = (1+1)(1+1)\prod(c_i+1) = 12

, which simplifies to

4\prod(c_i+1) = 12

, so

\prod(c_i+1) = 3

. Since

3

is prime, there is exactly one additional prime factor

p

with exponent

c = 2

Verify with the divisor count of

n^2

d(n^2) = (2(1)+1)(2(1)+1)(2(2)+1) = 3 \cdot 3 \cdot 5 = 45

Squaring

n

doubles all prime exponents, so

n^2 = 2^2 \cdot 3^2 \cdot p^4

, which has

(2+1)(2+1)(4+1) = 45

positive divisors, consistent with the given information.

Calculate the number of divisors of

n^2

that are multiples of

n

12

Any divisor of

n^2

has the form

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

with

0 \le x \le 2

0 \le y \le 2

, and

0 \le z \le 4

. For this divisor to be a multiple of

n = 2^1 \cdot 3^1 \cdot p^2

, the exponents must satisfy

1 \le x \le 2

(

2

choices),

1 \le y \le 2

(

2

choices), and

2 \le z \le 4

(

3

choices). Multiplying the choices gives

2 \cdot 2 \cdot 3 = 12

Anahtar Kavram

Divisor counting formula and prime factor exponent constraints

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