Two positive integers x and y satisfy x < y, \gcd(x, y) = 18, and \text{lcm}(x, y) = 1080.… — GRE General Test

Two positive integers $x$ and $y$ satisfy $x < y$ , $\gcd(x, y) = 18$ , and $\text{lcm}(x, y) = 1080$ . If $x$ has exactly $6$ positive divisors, what is the value of $y$ ?

A
$216$
B
$270$
C
$360$
D
$540$
$1080$ Answer

Answer

1080

Since

\gcd(x, y) = 18

and

\text{lcm}(x, y) = 1080

, we can express

x = 18m

and

y = 18n

with

\gcd(m, n) = 1

and

m < n

. Substituting into

\text{lcm}(x, y) = 18mn = 1080

gives

mn = 60

. The coprime pairs

(m, n)

with

m < n

are

(1, 60)

(3, 20)

(4, 15)

, and

(5, 12)

. Testing

x = 18m

for each pair: for

m = 1

x = 18 = 2^1 \cdot 3^2

, which has

(1+1)(2+1) = 6

positive divisors. This matches the condition, so

n = 60

and

y = 18 \times 60 = 1080

Step-by-Step Solution

Express

x

and

y

in terms of their Greatest Common Divisor

Let

x = 18m

and

y = 18n

, where

m

and

n

are positive integers such that

m < n

and

\gcd(m, n) = 1

Since

\gcd(x, y) = 18

, both numbers must be multiples of

18

, and their remaining factor parts must be coprime.

Use the LCM formula

\text{lcm}(x, y) = 18mn

to find

mn

18mn = 1080 \implies mn = 60

The least common multiple of

18m

and

18n

with

\gcd(m,n)=1

18mn

Find all coprime factor pairs

(m, n)

60

with

m < n

The prime factorization of

60

2^2 \cdot 3^1 \cdot 5^1

. Coprime pairs

(m, n)

are

(1, 60)

(3, 20)

(4, 15)

, and

(5, 12)

Pairs must be coprime so that

\gcd(18m, 18n) = 18

remains true.

Determine the number of positive divisors for

x = 18m

for each candidate pair

For

m = 1

x = 18 = 2^1 \cdot 3^2

, which has

(1+1)(2+1) = 6

positive divisors.

The number of positive divisors of

p_1^{a_1} p_2^{a_2} \dots

is given by

(a_1 + 1)(a_2 + 1) \dots

. Only

m = 1

gives exactly

6

divisors.

Calculate

y

y = 18n = 18 \cdot 60 = 1080

Corresponding to

m = 1

n = 60

gives

y = 1080

Key Concept

Prime Factorization, GCD/LCM Coprime Relationships, and Divisor Counting Formula

Alternative Method

Alternatively, use

\gcd(x, y) \cdot \text{lcm}(x, y) = x \cdot y \implies x \cdot y = 18 \cdot 1080 = 19440

. Since

x

has

6

positive divisors and

x

is a multiple of

18

x

must equal

18

because

18 = 2^1 \cdot 3^2

has

(1+1)(2+1) = 6

divisors. Then

y = 19440 / 18 = 1080

Estimated Time:1m 30s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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