If K = 2^a \times 3^b \times 5^c, where a, b, and c are positive integers, such that 15K h… — GMAT

If $K = 2^a \times 3^b \times 5^c$ , where $a$ , $b$ , and $c$ are positive integers, such that $15K$ has 15 more positive factors than $K$ , and $6K$ has 12 more positive factors than $K$ , what is the total number of positive factors of $K$ ?

A
10
12Answer
C
15
D
18
E
27

Answer

The total number of positive factors of

K

is 12.

By setting up factor count equations from prime exponents, we find

a=2, b=1, c=1

. Applying the formula

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

yields

(3)(2)(2) = 12

Step-by-Step Solution

Set up the formula for the number of positive factors of

K

15K

, and

6K

For

K = 2^a \cdot 3^b \cdot 5^c

, the number of factors is

d(K) = (a+1)(b+1)(c+1)

. For

15K = 2^a \cdot 3^{b+1} \cdot 5^{c+1}

d(15K) = (a+1)(b+2)(c+2)

. For

6K = 2^{a+1} \cdot 3^{b+1} \cdot 5^c

d(6K) = (a+2)(b+2)(c+1)

Prime factorizations determine the number of positive factors via the product of (exponent + 1).

Express the given factor differences algebraically.

d(15K) - d(K) = (a+1)[(b+2)(c+2) - (b+1)(c+1)] = (a+1)(b+c+3) = 15

. Also,

d(6K) - d(K) = (c+1)[(a+2)(b+2) - (a+1)(b+1)] = (c+1)(a+b+3) = 12

Factor out common terms to simplify the system of equations.

Solve for the positive integers

a

b

, and

c

Since

b \ge 1

and

c \ge 1

b+c+3 \ge 5

. The only factor of 15 greater than or equal to 5 is 5 (since

b+c+3=15

would require

a=0

). Thus

b+c+3=5 \implies b=1, c=1

and

a+1=3 \implies a=2

. Checking in the second equation:

(1+1)(2+1+3) = 2 \cdot 6 = 12

, which holds.

Integer constraints limit the potential factor pairs of 15.

Calculate

d(K)

d(K) = (2+1)(1+1)(1+1) = 3 \cdot 2 \cdot 2 = 12

Substitute

a=2, b=1, c=1

into the factor counting formula.

Key Concept

Prime Factorization and Number of Divisors

Estimated Time:2m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic