A positive integer n is divisible by 20 but is not divisible by 8. If n has exactly 15 pos… — GMAT

A positive integer $n$ is divisible by $20$ but is not divisible by $8$ . If $n$ has exactly $15$ positive divisors and $10n$ has exactly $24$ positive divisors, what is the value of $n$ ?

Cevap: 2500

Cevap

2500

To determine

n

, analyze its prime factorization. Divisibility by

20 = 2^2 \cdot 5^1

requires that

n

contains

2

raised to at least the power of

2

and

5

raised to at least the power of

1

. The condition that

n

is not divisible by

8 = 2^3

restricts the exponent of

2

to exactly

2

. Thus,

n = 2^2 \cdot 5^b \cdot K'

, where

b \ge 1

and

K'

is a product of powers of distinct prime factors other than

2

and

5

. The number of positive divisors of

n

d(n) = (2+1)(b+1)K = 3(b+1)K = 15

, which simplifies to

(b+1)K = 5

. Because

b \ge 1

, we have

b+1 \ge 2

. Since

5

is a prime number, its only divisor greater than or equal to

2

5

. Thus,

b+1 = 5

, giving

b = 4

, and

K = 1

, which means

n

has no prime factors other than

2

and

5

. Therefore,

n = 2^2 \cdot 5^4 = 4 \cdot 625 = 2500

. We verify that

10n = 2^3 \cdot 5^5

has

(3+1)(5+1) = 24

positive divisors, confirming the solution.

Adım Adım Çözüm

Determine the power of 2 in the prime factorization of

n

The exponent of

2

n

is exactly

2

Because

n

is a multiple of

20 = 2^2 \cdot 5

but not a multiple of

8 = 2^3

2^2

divides

n

but

2^3

does not.

Set up the divisor counting formula for

n

d(n) = (2+1)(b+1)K = 15 \implies (b+1)K = 5

, where

b \ge 1

is the exponent of

5

and

K

represents the product of terms from any additional prime factors

The total number of positive divisors of an integer

N = p_1^{a_1} p_2^{a_2} \dots

is given by

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

Solve for the exponents and prime factors of

n

b = 4

and

K = 1

, giving

n = 2^2 \cdot 5^4

Since

n

is divisible by

20

5

is a prime factor of

n

, so

b \ge 1

, which implies

b+1 \ge 2

. Since

5

is prime, its only factor greater than

1

5

, forcing

b+1 = 5

and

K = 1

Verify

d(10n)

and evaluate

n

10n = 2^3 \cdot 5^5

has

(3+1)(5+1) = 24

divisors, and

n = 4 \cdot 625 = 2500

Multiplying

n

10 = 2 \cdot 5

increases the exponent of

2

from

2

3

and the exponent of

5

from

4

5

Anahtar Kavram

Divisor Count Formula and Prime Factorization Constraints

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