Let n be a positive integer of the form n = 2^a \cdot 3^b, where a and b are positive inte… — GRE General Test

Let $n$ be a positive integer of the form $n = 2^a \cdot 3^b$ , where $a$ and $b$ are positive integers. If $n$ is divisible by $6$ and $n^2$ has exactly $35$ positive divisors, how many positive divisors does $6n$ have?

A
12
B
18
20Cevap
D
24
E
70

Cevap

The integer

6n

has 20 positive divisors.

The correct answer is 20. Since

n = 2^a \cdot 3^b

, we have

n^2 = 2^{2a} \cdot 3^{2b}

. The number of divisors of

n^2

(2a+1)(2b+1) = 35

. Given

n

is divisible by 6,

a, b \ge 1

, which means

2a+1 \ge 3

and

2b+1 \ge 3

. The factorizations of 35 are

5 \times 7

and

7 \times 5

, giving exponent values of 2 and 3. Then

6n = 2^{a+1} \cdot 3^{b+1}

, which has

(a+2)(b+2) = (2+2)(3+2) = 20

positive divisors.

Adım Adım Çözüm

Express

n^2

in terms of its prime factorization.

n^2 = (2^a \cdot 3^b)^2 = 2^{2a} \cdot 3^{2b}

Squaring an integer doubles the exponents of its prime factors.

Set up the equation for the number of positive divisors of

n^2

(2a + 1)(2b + 1) = 35

The number of positive divisors of

p_1^{e_1} p_2^{e_2}

is given by

(e_1 + 1)(e_2 + 1)

Solve for the exponents

a

and

b

given that

n

is divisible by 6.

Since

n

is divisible by 6, both

a \ge 1

and

b \ge 1

. Thus

2a + 1 > 1

and

2b + 1 > 1

. The only factor pairs of 35 into two integers greater than 1 are

5 \times 7

7 \times 5

. Therefore,

\{2a+1, 2b+1\} = \{5, 7\}

, which yields

\{a, b\} = \{2, 3\}

Divisibility by 6 guarantees both prime factors 2 and 3 are present in

n

Find the prime factorization of

6n

and calculate its number of positive divisors.

6n = (2^1 \cdot 3^1) \cdot (2^a \cdot 3^b) = 2^{a+1} \cdot 3^{b+1}

. Using

\{a, b\} = \{2, 3\}

, the exponents are

a+1

and

b+1

, which equal 3 and 4 (in either order). The number of divisors is

(3+1)(4+1) = 4 \times 5 = 20

Adding 1 to each exponent in

n

gives the exponents for

6n

, leading to

(a+2)(b+2) = (2+2)(3+2) = 20

Anahtar Kavram

Divisor Count Formula and Prime Factorization Constraints

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