If n is a positive integer with the prime factorization n = 2^a \times 3^b \times 5^c, whe… — GMAT

If $n$ is a positive integer with the prime factorization $n = 2^a \times 3^b \times 5^c$ , where $a$ , $b$ , and $c$ are positive integers, and $n$ has exactly $40$ positive integer divisors, what is the minimum possible value of $a + b + c$ ?

A
7
8Answer
C
9
D
11
E
14

Answer

The number of positive integer divisors of a number with prime factorization

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

is given by

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

. Setting this product equal to 40 and factoring 40 into three integers greater than 1 yields two distinct sets of factors:

\{2, 2, 10\}

and

\{2, 4, 5\}

. The sum of the factors in

\{2, 4, 5\}

2 + 4 + 5 = 11

, which is smaller than

2 + 2 + 10 = 14

. Subtracting 1 from each factor gives the exponents

a, b, c

1, 3, 4

, making the minimum sum

a+b+c = 1 + 3 + 4 = 8

Step-by-Step Solution

Express the number of divisors of

n

in terms of its exponents.

For

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

, the total number of positive divisors is

(a+1)(b+1)(c+1) = 40

The total divisor count formula requires taking the product of each prime factor's exponent increased by 1.

Find all combinations of three integers greater than 1 whose product is 40.

Since

a, b, c \ge 1

, we have

a+1, b+1, c+1 \ge 2

. The integer factorizations of 40 into 3 factors

\ge 2

are

(2, 2, 10)

and

(2, 4, 5)

We must break down 40 into 3 integer components corresponding to

(a+1)

(b+1)

, and

(c+1)

Calculate the sum

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

for each combination to find the minimum sum.

For

(2, 2, 10)

, the sum is

2 + 2 + 10 = 14

. For

(2, 4, 5)

, the sum is

2 + 4 + 5 = 11

. The minimum sum of factors is 11.

Minimizing

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

directly minimizes

a+b+c

Subtract 3 to obtain the minimum possible value of

a+b+c

a+b+c = 11 - 3 = 8

Since

(a+1)+(b+1)+(c+1) = (a+b+c) + 3 = 11

, subtracting 3 yields

a+b+c = 8

Key Concept

Divisor Count Formula from Prime Factorization

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