For a positive integer N = 2^a \times 3^b \times 5^c, where a, b, and c are positive integ… — GMAT

For a positive integer $N = 2^a \times 3^b \times 5^c$ , where $a$ , $b$ , and $c$ are positive integers, the integer $\frac{N}{2}$ has $40$ positive factors, the integer $\frac{N}{3}$ has $36$ positive factors, and the integer $\frac{N}{5}$ has $24$ positive factors. What is the total number of positive factors of $N^2$ ?

A
96
B
210
231Cevap
D
462
E
576

Cevap

231 positive factors

The correct answer is 231. By representing the factor counts of

\frac{N}{2}

\frac{N}{3}

, and

\frac{N}{5}

algebraically, we establish a system of equations for the exponents

a, b, c

. Solving this system yields

a=5

b=3

, and

c=1

. Squaring

N

doubles each exponent, giving

N^2 = 2^{10} \times 3^6 \times 5^2

. Applying the factor count formula gives

(10+1)(6+1)(2+1) = 231

Adım Adım Çözüm

Set up equations for the number of positive factors of N/2, N/3, and N/5 using the prime factorization formula.

The total number of factors of an integer

2^x 3^y 5^z

(x+1)(y+1)(z+1)

. Thus:

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

(a+1)b(c+1) = 36

, and

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

Dividing

N

by a prime factor reduces that prime factor's exponent by 1.

Express each equation in terms of

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

, the total number of factors of

N

(b+1)(c+1) = T - 40

(a+1)(c+1) = T - 36

, and

(a+1)(b+1) = T - 24

Expanding

(x-1)yz = xyz - yz = T - yz

allows expressing pairwise products in terms of

T

Multiply the three pairwise product equations to solve for

T

[(a+1)(b+1)(c+1)]^2 = (T-40)(T-36)(T-24) \implies T^2 = (T-40)(T-36)(T-24)

. Testing

T = 48

48^2 = 2304

and

(8)(12)(24) = 2304

. Thus

T = 48

Multiplying

(b+1)(c+1) \times (a+1)(c+1) \times (a+1)(b+1)

yields

T^2

Determine the individual exponents

a

b

, and

c

a+1 = \sqrt{\frac{(T-36)(T-24)}{T-40}} = \sqrt{\frac{12 \times 24}{8}} = 6 \implies a = 5

. Similarly,

b+1 = 4 \implies b = 3

, and

c+1 = 2 \implies c = 1

Dividing the product of two pairwise terms by the third gives the square of a single term.

Calculate the total number of positive factors of

N^2

N^2 = 2^{2a} \times 3^{2b} \times 5^{2c} = 2^{10} \times 3^6 \times 5^2

. Number of factors

= (10+1)(6+1)(2+1) = 11 \times 7 \times 3 = 231

Squaring

N

doubles each prime factor's exponent.

Anahtar Kavram

Divisibility, Prime Factorization, and Total Positive Factor Counting Formula

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