What is the smallest positive integer n such that n is a multiple of 180, the only prime f… — GRE General Test

What is the smallest positive integer $n$ such that $n$ is a multiple of $180$ , the only prime factors of $n$ are $2$ , $3$ , and $5$ , and $n$ has exactly $36$ positive integer divisors?

Answer: 1440

Answer

1440

To minimize

n = 2^a \cdot 3^b \cdot 5^c

subject to

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

with

a \ge 2

b \ge 2

, and

c \ge 1

, we evaluate all valid factor partitions of

36

. The partition

(a+1, b+1, c+1) = (6, 3, 2)

yields

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

, giving

n = 2^5 \cdot 3^2 \cdot 5^1 = 1440

, which is the smallest possible integer satisfying all criteria.

Step-by-Step Solution

Express

n

in terms of prime factorization and establish exponent inequalities.

n = 2^a \cdot 3^b \cdot 5^c

with

a \ge 2

b \ge 2

, and

c \ge 1

Since

n

is a multiple of

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

and contains no other prime factors, its prime powers must at least match those of

180

Set up the divisor count equation.

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

, where

a+1 \ge 3

b+1 \ge 3

, and

c+1 \ge 2

The total number of positive divisors of

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

is given by

(a+1)(b+1)(c+1)

Determine all valid factorizations of

36

into three factors

(x, y, z) = (a+1, b+1, c+1)

The valid factor sets

\{x, y, z\}

satisfying

x, y \ge 3

and

z \ge 2

are

\{4, 3, 3\}

and

\{6, 3, 2\}

Factor sets containing a factor of

2

for

x

y

(such as

\{9, 2, 2\}

) are invalid because

a+1 \ge 3

and

b+1 \ge 3

Calculate the value of

n

for all valid assignments of exponents.

From

\{4, 3, 3\}

(3, 2, 2) \implies 1800

(2, 3, 2) \implies 2700

(2, 2, 3) \implies 4500

.
From

\{6, 3, 2\}

(5, 2, 1) \implies 1440

(2, 5, 1) \implies 4860

Assigning larger exponents to smaller prime bases minimizes the overall product.

Select the minimum integer value among all candidates.

The smallest value is

1440

Comparing all valid candidates

1440, 1800, 2700, 4500, 4860

, the minimum is

1440

Key Concept

Prime Factorization and Divisor Count Constraints

Estimated Time:2m 30s

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