Two positive integers x and y have a greatest common divisor (GCD) of 12 and a least commo… — GRE General Test

Two positive integers $x$ and $y$ have a greatest common divisor (GCD) of $12$ and a least common multiple (LCM) of $360$ . Which of the following could be the value of $x$ ? Indicate all such values.

$24$ Answer
B
$48$
$60$ Answer
D
$90$
$120$ Answer

Answer

The possible values for

x

are

24

60

, and

120

Any valid value of

x

must be a multiple of

\gcd(x, y) = 12

and a divisor of

\text{lcm}(x, y) = 360

. Expressing

12

2^2 \cdot 3^1

and

360

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

x

must be of the form

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

with

2 \leq a \leq 3

1 \leq b \leq 2

, and

0 \leq c \leq 1

. The values

24

60

, and

120

meet all exponent constraints.

Step-by-Step Solution

Find the prime factorizations of the given GCD and LCM.

\gcd(x, y) = 12 = 2^2 \cdot 3^1 \cdot 5^0

and

\text{lcm}(x, y) = 360 = 2^3 \cdot 3^2 \cdot 5^1

Prime factorization allows analysis of exponent constraints for each prime factor.

Determine the constraints on any valid integer

x

Any valid value of

x

must be a multiple of

12

and a factor of

360

. Specifically,

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

where

2 \leq a \leq 3

1 \leq b \leq 2

, and

0 \leq c \leq 1

The GCD defines the minimum exponent for each prime factor, and the LCM defines the maximum exponent.

Test each provided option against the prime exponent bounds.

24 = 2^3 \cdot 3^1

satisfies the bounds (

a=3, b=1, c=0

48 = 2^4 \cdot 3^1

violates

a \leq 3

60 = 2^2 \cdot 3^1 \cdot 5^1

satisfies the bounds (

a=2, b=1, c=1

90 = 2^1 \cdot 3^2 \cdot 5^1

violates

a \geq 2

120 = 2^3 \cdot 3^1 \cdot 5^1

satisfies the bounds (

a=3, b=1, c=1

Options satisfying all exponent inequalities are valid possible values of

x

Key Concept

Prime exponent properties of GCD and LCM

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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