Let K = 2^5 \times 3^4 \times 5^3 \times 11^2. How many positive integer factors of K are … — GRE General Test

Let $K = 2^5 \times 3^4 \times 5^3 \times 11^2$ . How many positive integer factors of $K$ are divisible by $300$ but are not divisible by $900$ ?

A
16
24Cevap
C
36
D
48
E
96

Cevap

The correct answer 24 is obtained by analyzing the prime factorizations

300 = 2^2 \times 3^1 \times 5^2

and

900 = 2^2 \times 3^2 \times 5^2

. Any factor

f = 2^a \times 3^b \times 5^c \times 11^d

must satisfy

a \in \{2, 3, 4, 5\}

(4 choices),

b = 1

(1 choice, since

b \ge 1

for 300 but

b < 2

for 900),

c \in \{2, 3\}

(2 choices), and

d \in \{0, 1, 2\}

(3 choices). Multiplying these choices gives

4 \times 1 \times 2 \times 3 = 24

Adım Adım Çözüm

Express any factor of

K

in terms of prime factor exponent constraints.

Any factor

f

K

has the form

f = 2^a \times 3^b \times 5^c \times 11^d

, where

0 \le a \le 5

0 \le b \le 4

0 \le c \le 3

, and

0 \le d \le 2

The prime factors of

f

must be subsets of the prime factors of

K

with exponents not exceeding those in

K

Find the prime factorizations of

300

and

900

300 = 2^2 \times 3^1 \times 5^2

and

900 = 2^2 \times 3^2 \times 5^2

Divisibility criteria correspond to minimum exponent requirements for each prime factor.

Apply the divisibility conditions to determine constraints on each exponent.

For

f

to be divisible by

300

, we need

a \ge 2

b \ge 1

, and

c \ge 2

. For

f

to NOT be divisible by

900

, we must have

b < 2

. Thus,

b = 1

Combining

b \ge 1

and

b < 2

uniquely restricts

b

1

Count the number of valid choices for each exponent.

a \in \{2, 3, 4, 5\}

(

4

choices),

b \in \{1\}

(

1

choice),

c \in \{2, 3\}

(

2

choices), and

d \in \{0, 1, 2\}

(

3

choices).

The exponent

d

is unrestricted by

300

900

, so it can take any valid power present in

K

Multiply the number of independent choices using the fundamental counting principle.

4 \times 1 \times 2 \times 3 = 24

Each exponent choice can be paired independently to form a unique factor.

Anahtar Kavram

Counting Divisors Using Prime Exponent Constraints

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