Let N = 2^x \times 3^y \times 7^z, where x, y, and z are positive integers. If N is a mult… — GMAT

Let $N = 2^x \times 3^y \times 7^z$ , where $x$ , $y$ , and $z$ are positive integers. If $N$ is a multiple of $84$ and $N^2$ has exactly $45$ positive integer divisors, what is the value of $x + y + z$ ?

A
3
4Answer
C
5
D
6
E
7

Answer

The prime factorization of 84 is

2^2 \times 3^1 \times 7^1

. Because

N

is a multiple of 84, we must have

x \ge 2

y \ge 1

, and

z \ge 1

. The number of positive divisors of

N^2 = 2^{2x} \times 3^{2y} \times 7^{2z}

(2x+1)(2y+1)(2z+1) = 45

. Given

x \ge 2

2x+1 \ge 5

, while

2y+1 \ge 3

and

2z+1 \ge 3

. The only set of three factors of 45 meeting these criteria is

\{5, 3, 3\}

, giving

x=2, y=1, z=1

. The sum is

2 + 1 + 1 = 4

Step-by-Step Solution

Find the prime factorization of 84 to set lower bounds on

x, y, z

84 = 2^2 \times 3^1 \times 7^1

, so

x \ge 2

y \ge 1

, and

z \ge 1

Since

N

is a multiple of 84, its prime factorization must contain at least the prime factors of 84 with at least the same exponents.

Express the number of divisors of

N^2

N^2 = 2^{2x} \times 3^{2y} \times 7^{2z}

, so the number of positive divisors is

(2x + 1)(2y + 1)(2z + 1) = 45

Squaring a number doubles all its prime exponents, and the total number of divisors of

p_1^{e_1} p_2^{e_2} \dots

(e_1 + 1)(e_2 + 1)\dots

Determine the unique valid integer solution for

(2x+1)

(2y+1)

, and

(2z+1)

Since

x \ge 2

2x + 1 \ge 5

. Since

y, z \ge 1

2y+1 \ge 3

and

2z+1 \ge 3

. The factor triples of 45 into three odd factors

\ge 3

is uniquely

5 \times 3 \times 3

Factoring 45 gives

45 = 5 \times 3 \times 3

as the only breakdown satisfying

2x+1 \ge 5

Solve for

x, y, z

and compute their sum.

2x + 1 = 5 \implies x = 2

2y + 1 = 3 \implies y = 1

, and

2z + 1 = 3 \implies z = 1

. Thus

x + y + z = 2 + 1 + 1 = 4

Equating individual factors yields the exact exponent values.

Key Concept

Prime Factorization and Divisors of Powers of Integers

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