Let N = 2^a \cdot 3^b \cdot 5^c, where a, b, and c are positive integers. If N^2 has 105 p… — GMAT

Let $N = 2^a \cdot 3^b \cdot 5^c$ , where $a$ , $b$ , and $c$ are positive integers. If $N^2$ has $105$ positive divisors and $\dfrac{N}{2}$ has $16$ positive divisors, what is the number of positive divisors of $\dfrac{N}{15}$ ?

A
$8$
$9$ Cevap
C
$12$
D
$16$
E
$18$

Cevap

The number of positive divisors of

\dfrac{N}{15}

9

The prime factorization of

N^2

2^{2a} \cdot 3^{2b} \cdot 5^{2c}

, giving

(2a+1)(2b+1)(2c+1) = 105 = 3 \times 5 \times 7

. This implies that the exponents

a, b, c

are a permutation of

1, 2, 3

. The number of positive divisors of

\dfrac{N}{2} = 2^{a-1} \cdot 3^b \cdot 5^c

a(b+1)(c+1) = 16

. Testing

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

shows that only

a = 2

satisfies

2(b+1)(c+1) = 16

with

\{b, c\} = \{1, 3\}

. Finally,

\dfrac{N}{15} = 2^a \cdot 3^{b-1} \cdot 5^{c-1}

has

(a+1)bc

positive divisors. Substituting

a=2

and

bc = 3

gives

(2+1) \times 3 = 9

, making

9

the correct choice.

Adım Adım Çözüm

Express the number of positive divisors of

N^2

in terms of

a

b

, and

c

(2a+1)(2b+1)(2c+1) = 105

For a prime factorization

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

, the total number of positive divisors is

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

. Here

N^2 = 2^{2a} \cdot 3^{2b} \cdot 5^{2c}

Determine the set of values for the exponents

\{a, b, c\}

\{a, b, c\} = \{1, 2, 3\}

The prime factorization of

105

into three factors greater than

1

is uniquely

3 \times 5 \times 7

. Thus, the set of values for

\{2a+1, 2b+1, 2c+1\}

\{3, 5, 7\}

, which gives

\{2a, 2b, 2c\} = \{2, 4, 6\}

, so

\{a, b, c\} = \{1, 2, 3\}

Use the divisor count of

\dfrac{N}{2}

to identify the specific value of

a

a = 2

and

\{b, c\} = \{1, 3\}

Since

\dfrac{N}{2} = 2^{a-1} \cdot 3^b \cdot 5^c

, its divisor count is

a(b+1)(c+1) = 16

. Testing values from

\{1, 2, 3\}

for

a

: if

a=2

, then

2(b+1)(c+1) = 16 \implies (b+1)(c+1) = 8

. Since

\{b, c\} = \{1, 3\}

(1+1)(3+1) = 8

, which confirms

a = 2

and

\{b, c\} = \{1, 3\}

Calculate the number of positive divisors of

\dfrac{N}{15}

(2+1)(1)(3) = 9

\dfrac{N}{15} = \dfrac{N}{3 \cdot 5} = 2^a \cdot 3^{b-1} \cdot 5^{c-1}

. The number of positive divisors is

(a+1)(b-1+1)(c-1+1) = (a+1)bc

. Substituting

a=2

and

bc = 1 \times 3 = 3

yields

(2+1) \times 3 = 9

Anahtar Kavram

Divisor Count Formula for Prime Factorized Integers

Tahmini Süre:2m 0s

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