A positive integer N has a prime factorization of the form p^2 \times q, where p and q are… — ACT

A positive integer $N$ has a prime factorization of the form $p^2 \times q$ , where $p$ and $q$ are distinct prime numbers. If the sum of all the positive factors of $N$ (including $1$ and $N$ ) is $78$ , what is the value of $N$ ?

Answer: 45

Answer

The value of

N

is 45.

By applying the sum of divisors formula, the sum of factors of

N = p^2 \times q

is represented as

(1 + p + p^2)(1 + q) = 78

. Factoring 78 into two integers that satisfy the prime constraints of

p \ge 2

and

q \ge 2

yields

13 \times 6 = 78

. Solving

1 + p + p^2 = 13

gives

p = 3

, and

1 + q = 6

gives

q = 5

. Since 3 and 5 are distinct primes,

N = 3^2 \times 5 = 45

Step-by-Step Solution

Express the sum of factors of

N

algebraically

(1 + p + p^2)(1 + q) = 78

The sum of all positive factors of a number with prime factorization

p^a q^b

is given by the product of the sums of the powers of each prime factor.

Determine the constraints on

p

and

q

based on them being prime numbers

1 + p + p^2 \ge 7

and

1 + q \ge 3

The smallest prime number is 2, so

p \ge 2

and

q \ge 2

Find the factor pairs of 78 that satisfy the constraints

(1 + p + p^2, 1 + q) \in \{(13, 6), (26, 3)\}

The factors of 78 are 1, 2, 3, 6, 13, 26, 39, 78. We pair them such that one factor is at least 7 and the other is at least 3.

Solve for

p

and

q

for each possible factor pair

p = 3

and

q = 5

1 + p + p^2 = 13

, then

p^2 + p - 12 = 0

, which solves to

p = 3

(since

p > 0

). This leaves

1 + q = 6 \implies q = 5

. Both 3 and 5 are distinct primes. The other case,

1 + p + p^2 = 26

, has no integer solution for

p

Calculate the value of

N

N = 3^2 \times 5 = 45

Substitute the prime values back into the expression for

N

Key Concept

Sum of Divisors Formula and Prime Factorization

Estimated Time:1m 30s

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