Let n = 2^a \cdot 3^b \cdot 5, where a and b are positive integers. If n has a total of 24… — GMAT

Let $n = 2^a \cdot 3^b \cdot 5$ , where $a$ and $b$ are positive integers. If $n$ has a total of 24 positive divisors and has an equal number of even positive divisors and odd positive divisors, what is the value of $b$ ?

Answer: 5

Answer

For

n = 2^a \cdot 3^b \cdot 5^1

, the total number of divisors is

(a + 1)(b + 1)(2) = 24

, giving

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

. The odd divisors are formed strictly from

3^b \cdot 5^1

, giving

(b + 1)(2)

odd divisors. The even divisors require at least one factor of 2, giving

a(b + 1)(2)

even divisors. Setting even and odd divisor counts equal gives

2a(b + 1) = 2(b + 1)

, which reduces to

a = 1

. Substituting

a = 1

into

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

gives

2(b + 1) = 12

, leading directly to

b = 5

Step-by-Step Solution

Set up the formula for total positive divisors of

n

(a + 1)(b + 1)(1 + 1) = 24 \implies (a + 1)(b + 1) = 12

The total number of divisors of

p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}

(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)

Determine the counts of odd and even positive divisors.

Odd divisors =

2(b + 1)

, Even divisors =

2a(b + 1)

Odd divisors cannot contain any factors of 2 (exponent of 2 is 0). Even divisors must contain at least one factor of 2 (exponent of 2 can be

1, 2, \dots, a

Equate the number of odd and even divisors to find

a

2a(b + 1) = 2(b + 1) \implies a = 1

Dividing both sides by

2(b + 1)

(which is positive since

b \ge 1

) leaves

a = 1

Solve for

b

using the total divisor relation.

(1 + 1)(b + 1) = 12 \implies 2(b + 1) = 12 \implies b = 5

Substituting

a = 1

into

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

yields

b = 5

Key Concept

Counting total, odd, and even positive divisors using prime factorization exponents

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