For three positive integers a, b, and c, the greatest common divisor of any pair among the… — GMAT

For three positive integers $a$ , $b$ , and $c$ , the greatest common divisor of any pair among them is $12$ , and the least common multiple of all three integers is $5,040$ . If $a = 60$ , what is the minimum possible value of $b + c$ ?

A
$168$
$228$ Answer
C
$300$
D
$372$
E
$1,020$

Answer

The minimum possible value of

b + c

228

The value

228

is correct because prime factor analysis shows

b

and

c

must take exponents

\{4, 2\}

for prime 2,

\{2, 1\}

for prime 3,

\{0, 0\}

for prime 5, and

\{1, 0\}

for prime 7. Combining

2^4 \times 3^2 = 144

and

2^2 \times 3^1 \times 7^1 = 84

satisfies all pairwise GCD and LCM conditions while minimizing the sum to

144 + 84 = 228

Step-by-Step Solution

Express the given numbers and conditions in prime factorized form.

The pairwise GCD is

12 = 2^2 \times 3^1

. The LCM of

a, b, c

5,040 = 2^4 \times 3^2 \times 5^1 \times 7^1

. Given

a = 60 = 2^2 \times 3^1 \times 5^1

Prime factorization allows exact determination of the minimum and maximum required exponent for each prime factor across

a, b,

and

c

Determine the prime factor exponent constraints for

b

and

c

For prime factor 2: min exponent is 2, max is 4. Since

a

has

2^2

, exponents for

(b,c)

for factor 2 must be

\{4, 2\}

to minimize sum.
For prime factor 3: min exponent is 1, max is 2. Since

a

has

3^1

, exponents for

(b,c)

for factor 3 must be

\{2, 1\}

.
For prime factor 5: since

\gcd(a,b) = 12

and

\gcd(a,c) = 12

, 5 cannot divide

b

c

. Thus exponents for factor 5 are both 0.
For prime factor 7: max exponent is 1 in LCM, but

a

has

7^0

and pairwise GCD has

7^0

, so exactly one of

b

c

has

7^1

and the other has

7^0

The pairwise GCD dictates the minimum exponent present in all pairs, while the overall LCM dictates the maximum exponent present across the three numbers.

Test allocations of exponents to minimize

b + c

We must distribute exponents

\{4, 2\}

for 2,

\{2, 1\}

for 3, and

\{1, 0\}

for 7 between

b

and

c

.
Allocation 1:

b = 2^4 \times 3^2 = 144

and

c = 2^2 \times 3^1 \times 7^1 = 84 \implies b + c = 228

.
Allocation 2:

b = 2^4 \times 3^1 = 48

and

c = 2^2 \times 3^2 \times 7^1 = 252 \implies b + c = 300

.
Allocation 3:

b = 2^4 \times 3^1 \times 7^1 = 336

and

c = 2^2 \times 3^2 = 36 \implies b + c = 372

.
Allocation 4:

b = 2^4 \times 3^2 \times 7^1 = 1008

and

c = 2^2 \times 3^1 = 12 \implies b + c = 1020

Comparing all valid exponent combinations reveals the minimum sum.

Identify the minimum sum.

The minimum sum is

228

The smallest sum among all valid combinations is

144 + 84 = 228

Key Concept

Prime factor exponent extraction for pairwise GCD and joint LCM of three numbers

Estimated Time:2m 0s

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