A positive integer N is divisible by 10. The sum of all distinct prime factors of N is 10,… — State PSC Exam

A positive integer $N$ is divisible by $10$ . The sum of all distinct prime factors of $N$ is $10$ , and $N$ has exactly $12$ positive factors. What is the smallest possible value of $N$ ?

Answer: 60

Answer

The smallest possible value of

N

60

Because

N

is divisible by

10

, its prime factorization must contain

2

and

5

. The sum of all distinct prime factors is

10

, which requires

2 + 5 + p = 10

, giving

p = 3

. Hence,

N = 2^a \times 3^b \times 5^c

with

a, b, c \ge 1

. The number of factors is

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

. The unique product representation of

12

using three integers

\ge 2

3 \times 2 \times 2

, meaning the exponents are one

2

and two

1

s. Minimizing

N

requires assigning the highest exponent

2

to the smallest prime

2

, giving

N = 2^2 \times 3^1 \times 5^1 = 60

Step-by-Step Solution

Identify the distinct prime factors of

N

The distinct prime factors are

2, 3,

and

5

Divisibility by

10

implies

2

and

5

are prime factors of

N

. The sum of distinct prime factors is

10

, so the remaining prime factor is

10 - (2 + 5) = 3

Determine the exponents in the prime factorization of

N

The set of exponents

\{a, b, c\}

must be

\{2, 1, 1\}

The total number of factors is

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

. Since

a, b, c \ge 1

, the only integer partition of

12

into three factors

\ge 2

3 \times 2 \times 2

Assign exponents to minimize the value of

N

N = 2^2 \times 3^1 \times 5^1 = 60

To minimize

N

, the largest exponent

2

must be assigned to the smallest prime base

2

Key Concept

Prime factorization, distinct prime factor properties, and total factor count formula

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