Let N be a positive integer with exactly 12 positive divisors. If the sum of the distinct … — GRE General Test

Let $N$ be a positive integer with exactly $12$ positive divisors. If the sum of the distinct prime factors of $N$ is $12$ and $N$ is not divisible by $4$ , what is the least possible value of $N$ ?

Answer: 126

Answer

126

The least possible value of

N

is 126. The only sets of distinct prime factors summing to 12 are

\{5, 7\}

and

\{2, 3, 7\}

. Using

\{2, 3, 7\}

produces smaller candidates. Because

N

is not divisible by 4, the exponent of 2 must be 1. The total divisor condition

(1+1)(b+1)(c+1) = 12

requires

(b+1)(c+1) = 6

, giving exponent pairs

(1, 2)

(2, 1)

for bases 3 and 7. Assigning the exponent 2 to 3 and 1 to 7 minimizes

N

, giving

2^1 \cdot 3^2 \cdot 7^1 = 126

Step-by-Step Solution

Find all sets of distinct prime factors that sum to 12.

The possible sets of distinct prime factors are

\{5, 7\}

and

\{2, 3, 7\}

Testing combinations of prime numbers (

2, 3, 5, 7, 11, \dots

5 + 7 = 12

and

2 + 3 + 7 = 12

are the only valid sets of distinct primes summing to 12.

Analyze the set of prime factors

\{2, 3, 7\}

under the condition that

N

is not divisible by 4.

The prime factorization is

N = 2^a \cdot 3^b \cdot 7^c

, where

a = 1

Since 2 is a prime factor of

N

a \ge 1

. Because

N

is not divisible by 4 (

2^2

a

must be strictly less than 2. Thus,

a = 1

Determine the exponents

b

and

c

using the total number of positive divisors.

The number of divisors is

(1+1)(b+1)(c+1) = 12

, which simplifies to

(b+1)(c+1) = 6

. The possible integer pairs

(b, c)

for

b, c \ge 1

are

(1, 2)

and

(2, 1)

The divisor count formula for

N = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k}

(e_1+1)(e_2+1)\dots(e_k+1) = 12

Calculate values of

N

for these exponent pairs.

For

(b, c) = (1, 2)

N = 2^1 \cdot 3^1 \cdot 7^2 = 294

. For

(b, c) = (2, 1)

N = 2^1 \cdot 3^2 \cdot 7^1 = 126

To make

N

as small as possible, assign the larger exponent to the smaller prime base (

3^2 \cdot 7^1 < 3^1 \cdot 7^2

Compare with candidate values from the alternative set of prime factors

\{5, 7\}

For

\{5, 7\}

N = 5^3 \cdot 7^2 = 6125

5^5 \cdot 7^1 = 21875

, both of which are much larger than 126.

The combination

(a+1)(b+1) = 12

yields exponents of 3 and 2 (or 5 and 1), resulting in much higher prime powers.

Key Concept

Divisor count formula combined with prime factorization constraints

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