If n = 18^4 \times 35^3 \times 22^2, what is the total number of prime factors of n, count… — GMAT

If $n = 18^4 \times 35^3 \times 22^2$ , what is the total number of prime factors of $n$ , counting multiplicities (the sum of the exponents in its prime factorization)?

A
5
B
9
C
18
22Answer
E
23

Answer

The total number of prime factors of

n

, counting multiplicities, is 22.

The correct answer is obtained by rewriting each base into prime factors (

18 = 2 \times 3^2

35 = 5 \times 7

22 = 2 \times 11

), combining powers of equal bases to obtain

n = 2^6 \times 3^8 \times 5^3 \times 7^3 \times 11^2

, and summing the resulting exponents (

6 + 8 + 3 + 3 + 2 = 22

Step-by-Step Solution

Express each composite base in terms of its prime factors.

18 = 2 \times 3^2

35 = 5 \times 7

, and

22 = 2 \times 11

Prime factorization requires writing every base strictly as a product of prime numbers.

Substitute the prime factorizations into the expression for

n

and apply exponent rules.

n = (2 \times 3^2)^4 \times (5 \times 7)^3 \times (2 \times 11)^2 = (2^4 \times 3^8) \times (5^3 \times 7^3) \times (2^2 \times 11^2)

Distribute exponents over multiplication using

(a \times b)^k = a^k \times b^k

and

(a^m)^n = a^{m \cdot n}

Combine like prime bases by adding their exponents.

n = 2^{4+2} \times 3^8 \times 5^3 \times 7^3 \times 11^2 = 2^6 \times 3^8 \times 5^3 \times 7^3 \times 11^2

Combine terms with identical bases according to exponent rules (

a^m \times a^n = a^{m+n}

Sum the exponents of all prime factors to find the total count including multiplicities.

6 + 8 + 3 + 3 + 2 = 22

The total number of prime factors counted with multiplicity is given by the sum of exponents in the canonical prime factorization.

Key Concept

Prime Factorization and Exponent Rules

Estimated Time:1m 30s

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