Let m and n be positive integers whose prime factorizations consist only of the prime fact… — GRE General Test

Let $m$ and $n$ be positive integers whose prime factorizations consist only of the prime factors $2$ , $3$ , and $5$ . The greatest common divisor of $m$ and $n$ is $360$ , and their least common multiple is $108,000$ . If the power of $2$ in the prime factorization of $m$ is strictly greater than the power of $2$ in the prime factorization of $n$ , and $m$ has exactly $36$ positive divisors, what is the total number of positive divisors of $n$ ?

A
$32$
B
$48$
$64$ Cevap
D
$72$
E
$96$

Cevap

The total number of positive divisors of

n

64

By writing the GCD and LCM in prime factorized form,

\text{GCD}(m,n) = 2^3 \times 3^2 \times 5^1

and

\text{LCM}(m,n) = 2^5 \times 3^3 \times 5^3

. Given

a > x

for the exponent of

2

, we obtain

a=5

and

x=3

. Using the divisor count formula for

m

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

, which requires

(b+1)(c+1) = 6

. Testing possible exponent values

b \in \{2,3\}

reveals that only

b=2

yields an integer exponent

c=1

. This uniquely determines

n = 2^3 \times 3^3 \times 5^3

. Calculating

(3+1)(3+1)(3+1)

gives

64

positive divisors.

Adım Adım Çözüm

Express the greatest common divisor (GCD) and least common multiple (LCM) in prime factorization form.

\text{GCD}(m, n) = 360 = 2^3 \times 3^2 \times 5^1

and

\text{LCM}(m, n) = 108,000 = 2^5 \times 3^3 \times 5^3

Prime factorizations allow direct comparison of the minimum and maximum exponents for each prime factor.

Determine the exponent of

2

for both integers.

For prime factor

2

\min(a, x) = 3

and

\max(a, x) = 5

. Since the exponent of

2

m

is strictly greater than in

n

(

a > x

a = 5

and

x = 3

The GCD gives the lower exponent bound while the LCM gives the upper exponent bound.

Use the number of divisors of

m

to solve for its remaining exponents.

Let

m = 2^5 \times 3^b \times 5^c

. The number of positive divisors of

m

(5+1)(b+1)(c+1) = 6(b+1)(c+1) = 36

, so

(b+1)(c+1) = 6

. Since

b \in \{2, 3\}

, if

b = 3

then

b+1=4

which does not divide

6

. Thus,

b = 2

and

c = 1

The divisor count formula requires integer exponents, uniquely identifying

b = 2

and

c = 1

for

m

Determine the exponents for

n

and calculate its divisor count.

Since

b = 2

and

\max(b, y) = 3

y = 3

. Since

c = 1

and

\max(c, z) = 3

z = 3

. Thus

n = 2^3 \times 3^3 \times 5^3

, and its divisor count is

(3+1)(3+1)(3+1) = 64

Combining the max/min constraints from LCM/GCD yields all exponents of

n

Anahtar Kavram

GCD and LCM Prime Exponent Relations & Divisor Counting Formula

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