Three automated signal beacons emit flashes at regular time intervals of x, y, and z secon… — GMAT

Three automated signal beacons emit flashes at regular time intervals of $x$ , $y$ , and $z$ seconds, respectively, where $x$ , $y$ , and $z$ are positive integers. If $\text{gcd}(x, y) = 18$ , $\text{gcd}(y, z) = 24$ , and $\text{gcd}(x, z) = 30$ , what is the minimum possible value of $\text{lcm}(x, y, z)$ ?

A
$180$
$360$ Answer
C
$720$
D
$1{}080$
E
$12{}960$

Answer

360

By breaking each given greatest common divisor into its prime factorization, we establish minimum requirements for the exponents of prime factors

2

3

, and

5

across

x

y

, and

z

. Specifically,

\text{gcd}(y, z) = 24 = 2^3 \times 3

dictates that both

y

and

z

must contain

2^3

, making the maximum exponent of

2

equal to

3

. Similarly,

\text{gcd}(x, y) = 18 = 2 \times 3^2

forces the maximum exponent of

3

to be

2

, and

\text{gcd}(x, z) = 30 = 2 \times 3 \times 5

forces the maximum exponent of

5

to be

1

. Multiplying these prime powers together gives

2^3 \times 3^2 \times 5 = 360

Step-by-Step Solution

Express the given pairwise greatest common divisors in prime factorized form.

\text{gcd}(x, y) = 18 = 2^1 \times 3^2

\text{gcd}(y, z) = 24 = 2^3 \times 3^1

\text{gcd}(x, z) = 30 = 2^1 \times 3^1 \times 5^1

GCD conditions determine the minimum prime factor exponents across pairs of variables.

Analyze the exponents for prime factor

2

\min(v_2(x), v_2(y)) = 1

\min(v_2(y), v_2(z)) = 3

, and

\min(v_2(x), v_2(z)) = 1

. This implies

v_2(y) \ge 3

and

v_2(z) \ge 3

, while

v_2(x) = 1

. To minimize the LCM, set

v_2(x) = 1

v_2(y) = 3

v_2(z) = 3

. Thus,

\max(v_2(x), v_2(y), v_2(z)) = 3

The exponent of

2

\text{lcm}(x, y, z)

is the maximum of the individual exponents of

2

Analyze the exponents for prime factor

3

\min(v_3(x), v_3(y)) = 2

\min(v_3(y), v_3(z)) = 1

, and

\min(v_3(x), v_3(z)) = 1

. This implies

v_3(x) \ge 2

and

v_3(y) \ge 2

, while

v_3(z) = 1

. To minimize the LCM, set

v_3(x) = 2

v_3(y) = 2

v_3(z) = 1

. Thus,

\max(v_3(x), v_3(y), v_3(z)) = 2

The exponent of

3

\text{lcm}(x, y, z)

is the maximum of the individual exponents of

3

Analyze the exponents for prime factor

5

\min(v_5(x), v_5(y)) = 0

\min(v_5(y), v_5(z)) = 0

, and

\min(v_5(x), v_5(z)) = 1

. This requires

v_5(x) \ge 1

and

v_5(z) \ge 1

, while

v_5(y) = 0

. To minimize the LCM, set

v_5(x) = 1

v_5(y) = 0

v_5(z) = 1

. Thus,

\max(v_5(x), v_5(y), v_5(z)) = 1

The exponent of

5

\text{lcm}(x, y, z)

is the maximum of the individual exponents of

5

Calculate the minimum value of

\text{lcm}(x, y, z)

\text{lcm}(x, y, z) = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360

No additional prime factors are required to satisfy the conditions, yielding the minimum possible LCM.

Key Concept

Prime Exponent Constraints on Pairwise GCD and LCM

Estimated Time:2m 0s

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