For two positive integers x and y, their greatest common divisor is \gcd(x, y) = 12 and th… — GMAT

For two positive integers $x$ and $y$ , their greatest common divisor is $\gcd(x, y) = 12$ and their least common multiple is $\text{lcm}(x, y) = 10,800$ . If $x > y$ , $x$ is not divisible by $9$ , and $y$ is not divisible by $25$ , what is the value of $x + y$ ?

A
732
B
972
1,308Answer
D
1,356
E
1,500

Answer

The sum of the two integers

x + y

1,308

By prime factorizing the given GCD (

2^2 \cdot 3^1

) and LCM (

2^4 \cdot 3^3 \cdot 5^2

), the minimum and maximum exponents for each prime factor are determined. The constraint that

x

is not divisible by

9

forces the exponent of

3

x

to be

1

, so the exponent of

3

y

must be

3

. The constraint that

y

is not divisible by

25

forces the exponent of

5

y

to be

0

, so the exponent of

5

x

must be

2

. Finally, the condition

x > y

requires the exponent of

2

x

to be

4

and in

y

to be

2

. Thus,

x = 1,200

and

y = 108

, giving

x + y = 1,308

Step-by-Step Solution

Express the GCD and LCM in their prime factorizations.

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

and

\text{lcm}(x, y) = 10,800 = 2^4 \cdot 3^3 \cdot 5^2

Decomposing into prime factors allows direct determination of the minimum and maximum powers of each prime factor present in

x

and

y

Determine the prime exponents for

x

and

y

using the given divisibility constraints.

For prime factor

3

\min(b_1, b_2) = 1

and

\max(b_1, b_2) = 3

. Since

x

is not divisible by

9 = 3^2

b_1 = 1

, which forces

b_2 = 3

. For prime factor

5

\min(c_1, c_2) = 0

and

\max(c_1, c_2) = 2

. Since

y

is not divisible by

25 = 5^2

c_2 < 2

, which forces

c_1 = 2

and

c_2 = 0

The GCD gives the minimum exponent of each prime factor across both numbers, while the LCM gives the maximum exponent.

Apply the condition

x > y

to assign the powers of

2

The exponents for

2

must be

\{2, 4\}

. If

x

takes exponent

4

and

y

takes exponent

2

x = 2^4 \cdot 3^1 \cdot 5^2 = 1,200

and

y = 2^2 \cdot 3^3 \cdot 5^0 = 108

. Since

1,200 > 108

, this satisfies

x > y

x

took exponent

2

and

y

took exponent

4

, then

x = 300

and

y = 432

, violating

x > y

Calculate

x + y

x + y = 1,200 + 108 = 1,308

Adding the uniquely determined values of

x

and

y

gives the final requested sum.

Key Concept

Prime factor exponent min/max rules for GCD and LCM:

\gcd(x,y)

takes the minimum exponent of each prime factor, while

\text{lcm}(x,y)

takes the maximum exponent.

Estimated Time:2m 30s

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Step-by-Step Solution

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