Let k be a positive integer. The integer k has exactly 6 positive divisors, 3k has exactly… — GRE General Test

Let $k$ be a positive integer. The integer $k$ has exactly $6$ positive divisors, $3k$ has exactly $8$ positive divisors, and $5k$ has exactly $12$ positive divisors. Which of the following could be the value of $k$ ? Indicate all such values.

$18$ Cevap
B
$45$
$63$ Cevap
D
$81$
$99$ Cevap

Cevap

The correct values of

k

are

18

63

, and

99

The integers

18

63

, and

99

are all of the form

k = 3^2 \times q^1 = 9q

, where

q

is a prime number other than

3

5

(specifically

q = 2, 7, 11

). Each has

(2+1)(1+1) = 6

divisors,

3k = 3^3 \times q^1

has

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

divisors, and

5k = 3^2 \times 5^1 \times q^1

has

(2+1)(1+1)(1+1) = 12

divisors.

Adım Adım Çözüm

Analyze the prime factorization form of

k

based on its divisor count

d(k) = 6

Since

6 = 6 \times 1

3 \times 2

k

must be of the form

p^5

p^2 q^1

, where

p

and

q

are distinct prime numbers.

The formula for the total number of positive divisors of an integer

n = p_1^{a_1} p_2^{a_2} \dots p_m^{a_m}

(a_1 + 1)(a_2 + 1) \dots (a_m + 1)

Test the prime factorization form

k = p^5

k = p^5

is impossible.

p \neq 3

, then

3k = 3^1 p^5

, which would have

(1+1)(5+1) = 12

divisors (contradicting

d(3k) = 8

). If

p = 3

3k = 3^6

has

7

divisors. If

p = 5

5k = 5^6

has

7

divisors.

Test the prime factorization form

k = p^2 q^1

under different prime identities.

k

must be of the form

3^2 q^1

, where

q

is a prime number distinct from

3

and

5

p = 3

and

q \neq 3, 5

, then

3k = 3^3 q^1

has

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

divisors, and

5k = 3^2 \times 5^1 \times q^1

has

(2+1)(1+1)(1+1) = 12

divisors. Any other choice for

p

q

fails to yield

d(3k) = 8

d(5k) = 12

Evaluate the given options against the required form

k = 9q

where

q

is a prime distinct from

3

and

5

18 = 9 \times 2

(

q=2

, valid prime),

63 = 9 \times 7

(

q=7

, valid prime), and

99 = 9 \times 11

(

q=11

, valid prime) are correct.

45 = 9 \times 5

(

q=5

, invalid because

q \neq 5

) and

81 = 3^4

(invalid form) are incorrect.

Only primes

q \notin \{3, 5\}

preserve the required divisor counts for

3k

and

5k

Anahtar Kavram

Divisors formula and impact of prime multiplication on prime factorization exponents

Tahmini Süre:2m 30s

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru