Let M = 2^a \times 3^b \times 5^c and N = 2^c \times 3^a \times 5^b, where a, b, and c are… — GMAT

Let $M = 2^a \times 3^b \times 5^c$ and $N = 2^c \times 3^a \times 5^b$ , where $a$ , $b$ , and $c$ are distinct positive integers. If the greatest common divisor of $M$ and $N$ is $90$ and the least common multiple of $M$ and $N$ is $32,400$ , what is the value of $a + b + c$ ?

A
6
7Cevap
C
8
D
9
E
14

Cevap

The value of

a + b + c

is 7.

The correct answer is 7. By applying the fundamental property

M \times N = \gcd(M, N) \times \text{lcm}(M, N)

, we get

2^{a+c} \times 3^{a+b} \times 5^{b+c} = 2^5 \times 3^6 \times 5^3

. Matching exponents gives

a + c = 5

a + b = 6

, and

b + c = 3

. Summing these three equations yields

2(a + b + c) = 14

, so

a + b + c = 7

Adım Adım Çözüm

Express the GCD and LCM of

M

and

N

in terms of their prime factorizations.

\gcd(M,N) = 90 = 2^1 \times 3^2 \times 5^1

and

\text{lcm}(M,N) = 32,400 = 2^4 \times 3^4 \times 5^2

Prime factorization allows us to relate the exponents of

M

and

N

directly to their GCD and LCM.

Use the identity

M \times N = \gcd(M,N) \times \text{lcm}(M,N)

to multiply the two numbers.

(

2^a \times 3^b \times 5^c) \times (2^c \times 3^a \times 5^b) = (2^1 \times 3^2 \times 5^1) \times (2^4 \times 3^4 \times 5^2)

, which simplifies to

2^{a+c} \times 3^{a+b} \times 5^{b+c} = 2^{1+4} \times 3^{2+4} \times 5^{1+2} = 2^5 \times 3^6 \times 5^3

The product of two positive integers is equal to the product of their greatest common divisor and least common multiple.

Equate the exponents for each prime base

2

3

, and

5

a + c = 5

a + b = 6

, and

b + c = 3

Since prime bases are unique, exponents of corresponding prime factors on both sides of the equation must be equal.

Sum the three equations and solve for

a + b + c

(a + c) + (a + b) + (b + c) = 5 + 6 + 3 \implies 2(a + b + c) = 14 \implies a + b + c = 7

Adding the three system equations yields twice the desired sum.

Anahtar Kavram

Prime Factorization, GCD and LCM Product Relationship

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