A positive integer N has exactly 15 positive factors and is divisible by 6. If N has exact… — State PSC Exam

A positive integer $N$ has exactly $15$ positive factors and is divisible by $6$ . If $N$ has exactly two distinct prime factors, what is the sum of all possible values of $N$ that are less than $500$ ?

A
144
B
324
468Answer
D
540

Answer

The sum of all possible values of N less than 500 is 468.

For a number

N = p^a \times q^b

to have 15 factors,

(a+1)(b+1) = 15

. The factor pairs of 15 for two prime factors are

(5, 3)

and

(3, 5)

, which correspond to exponent pairs

(4, 2)

and

(2, 4)

. Since

N

is divisible by 6, its prime factors are 2 and 3. Computing both cases gives

2^4 \times 3^2 = 144

and

2^2 \times 3^4 = 324

. Both are less than 500, making their sum

144 + 324 = 468

Step-by-Step Solution

Determine the prime factors of N

N has prime factors 2 and 3

Since N is divisible by 6, it must have at least 2 and 3 as prime factors. The problem states N has exactly two distinct prime factors, so its prime factorization is of the form

N = 2^a \times 3^b

Apply the total factors formula

(a + 1)(b + 1) = 15

The total number of positive factors of

N = p^a \times q^b

is given by

(a + 1)(b + 1) = 15

Find valid non-negative integer pairs (a, b)

Either

(a=4, b=2)

(a=2, b=4)

Since 15 factors into

5 \times 3

3 \times 5

(note that

15 \times 1

would imply only one prime factor, which contradicts having two distinct prime factors), the possible exponent pairs are

(4, 2)

and

(2, 4)

Calculate the values of N and check bounds

N_1 = 2^4 \times 3^2 = 144

and

N_2 = 2^2 \times 3^4 = 324

Both

144 < 500

and

324 < 500

meet all conditions.

Sum the valid values of N

144 + 324 = 468

Adding the two valid integers yields the total sum.

Key Concept

Factors and Prime Factorization Exponent Rule

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