If N = 2^4 \times 3^3 \times 5^2, how many positive integer factors of N are divisible by … — GRE General Test

If $N = 2^4 \times 3^3 \times 5^2$ , how many positive integer factors of $N$ are divisible by $6$ but not divisible by $12$ ?

Cevap: 9

Cevap

The number of positive integer factors of

N

divisible by 6 but not by 12 is 9.

Any factor of

N = 2^4 \times 3^3 \times 5^2

is of the form

2^a \times 3^b \times 5^c

. For the factor to be divisible by

6

, we must have

a \ge 1

and

b \ge 1

. For it to not be divisible by

12 = 2^2 \times 3^1

, we must have

a < 2

. Therefore,

a

must equal

1

. The possible choices for

a

are

1

value (

1

), for

b

are

3

values (

1, 2, 3

), and for

c

are

3

values (

0, 1, 2

). Multiplying these options gives

1 \times 3 \times 3 = 9

Adım Adım Çözüm

Express the prime factorization structure of a factor of

N

Any factor of

N

takes the form

2^a \times 3^b \times 5^c

with bounds

0 \le a \le 4

0 \le b \le 3

, and

0 \le c \le 2

Divisors of a number are formed by taking prime factors with exponents between zero and their maximum powers in the original number.

Determine exponent constraints for divisibility by 6 and non-divisibility by 12.

Divisibility by

6

requires

a \ge 1

and

b \ge 1

. Non-divisibility by

12

requires

a < 2

. Thus,

a = 1

exactly.

A factor must contain at least one factor of 2 and one factor of 3 to be a multiple of 6, but containing two or more factors of 2 makes it a multiple of 12.

Count combinations of choices for the exponents.

Exponent

a

has

1

option (

a = 1

b

has

3

options (

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

), and

c

has

3

options (

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

). Total

= 1 \times 3 \times 3 = 9

Applying the fundamental counting principle by multiplying the number of choices for independent prime factor exponents.

Anahtar Kavram

Counting Divisors with Prime Factorization and Divisibility Constraints

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