Let n = 2^a \cdot 3^b \cdot p^c, where a, b, and c are positive integers and p is a prime … — GRE General Test

Let $n = 2^a \cdot 3^b \cdot p^c$ , where $a$ , $b$ , and $c$ are positive integers and $p$ is a prime number strictly greater than $5$ . The integer $n$ has exactly $36$ positive divisors, and $\gcd(n, 180) = 36$ . Which of the following statements MUST be true? Select all such statements.

The integer $n$ is divisible by $36$ .Answer
The exponent $c$ cannot exceed $3$ .Answer
C
The sum of the exponents $a + b + c$ must equal $7$ .
The least common multiple of $n$ and $180$ is equal to $5n$ .Answer
E
The exponent $a$ must equal $2$ .

Answer

The statements asserting that

n

is divisible by

36

, that the exponent

c

cannot exceed

3

, and that

\text{lcm}(n, 180) = 5n

must be true.

The statement that

n

is divisible by

36

is true because

a \ge 2

and

b \ge 2

. The statement that

c \le 3

is true because

(a+1)(b+1) \ge 9

, bounding

c+1 \le 4

. The statement that

\text{lcm}(n, 180) = 5n

is true because

\max(a, 2) = a

\max(b, 2) = b

, and

180

contributes a factor of

5^1

Step-by-Step Solution

Analyze the GCD condition to establish bounds on exponents

a

and

b

a \ge 2

and

b \ge 2

Since

180 = 2^2 \cdot 3^2 \cdot 5^1

and

\gcd(n, 180) = 36 = 2^2 \cdot 3^2

, the minimum of the exponents of

2

2

(so

a \ge 2

) and the minimum of the exponents of

3

2

(so

b \ge 2

Use the divisor count formula

(a+1)(b+1)(c+1) = 36

to find constraints on

c

c \le 3

Since

a \ge 2 \implies a+1 \ge 3

and

b \ge 2 \implies b+1 \ge 3

, the product

(a+1)(b+1) \ge 9

. Therefore,

c+1 = \frac{36}{(a+1)(b+1)} \le \frac{36}{9} = 4

, which gives

c \le 3

Evaluate the divisibility of

n

36

n

is divisible by

36

Because

a \ge 2

and

b \ge 2

2^2 \cdot 3^2 = 36

divides

2^a \cdot 3^b \cdot p^c = n

Calculate

\text{lcm}(n, 180)

\text{lcm}(n, 180) = 5n

\text{lcm}(2^a \cdot 3^b \cdot p^c, 2^2 \cdot 3^2 \cdot 5^1) = 2^{\max(a,2)} \cdot 3^{\max(b,2)} \cdot 5^1 \cdot p^c = 2^a \cdot 3^b \cdot 5 \cdot p^c = 5n

since

a \ge 2

b \ge 2

, and

p > 5

Test counterexamples for statement constancy regarding

a+b+c=7

and

a=2

Neither

a+b+c=7

nor

a=2

is required.

The choice

(a, b, c) = (5, 2, 1)

satisfies

(5+1)(2+1)(1+1) = 36

and

\gcd(n, 180) = 36

, yielding

a+b+c=8 \neq 7

and

a=5 \neq 2

Key Concept

Prime Factorization, Greatest Common Divisor (GCD), and Least Common Multiple (LCM) properties

Question

Answer

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Key Concept

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