If a and b are positive integers, is a^2b divisible by 12? (1) ab^2 is divisible by 18. (2… — GMAT

If $a$ and $b$ are positive integers, is $a^2b$ divisible by $12$ ?

(1) $ab^2$ is divisible by $18$ .
(2) $a^3b$ is divisible by $72$ .

A
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.Answer
C
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D
EACH statement ALONE is sufficient.
E
Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

Statement (2) alone guarantees that

a^3b

contains at least three factors of

2

and two factors of

3

. Analyzing the exponent bounds for non-negative integers proves that

a^2b

must contain at least two factors of

2

and one factor of

3

, ensuring

a^2b

is divisible by

12

. Statement (1) alone is insufficient because

a=3, b=6

makes

ab^2=108

(divisible by

18

) but

a^2b=54

(not divisible by

12

). Thus, Statement (2) ALONE is sufficient.

Step-by-Step Solution

Rephrase the target question in terms of prime factorization

For

a^2b

to be divisible by

12 = 2^2 \times 3^1

, we need

2 v_2(a) + v_2(b) \ge 2

and

2 v_3(a) + v_3(b) \ge 1

, where

v_p(n)

is the exponent of prime

p

in the prime factorization of

n

Decomposing divisibility into prime factor exponent inequalities allows definitive evaluation of sufficiency.

Evaluate Statement (1):

ab^2

is divisible by

18 = 2^1 \times 3^2

This gives

v_2(a) + 2 v_2(b) \ge 1

and

v_3(a) + 2 v_3(b) \ge 2

. Test counterexample

a=3, b=6

ab^2 = 3 \times 36 = 108

(divisible by 18), but

a^2b = 9 \times 6 = 54

, which is not divisible by 12.

A single valid counterexample proves Statement (1) is NOT sufficient.

Evaluate Statement (2):

a^3b

is divisible by

72 = 2^3 \times 3^2

This gives

3 v_2(a) + v_2(b) \ge 3

and

3 v_3(a) + v_3(b) \ge 2

. If

v_2(a) = 0

, then

v_2(b) \ge 3 \implies 2 v_2(a) + v_2(b) \ge 3 \ge 2

. If

v_2(a) \ge 1

, then

2 v_2(a) + v_2(b) \ge 2(1) + 0 = 2

. Similarly for prime 3: if

v_3(a) = 0

v_3(b) \ge 2 \implies 2 v_3(a) + v_3(b) \ge 2 \ge 1

; if

v_3(a) \ge 1

2 v_3(a) + v_3(b) \ge 2 \ge 1

. Thus

a^2b

is always divisible by 12.

Statement (2) strictly guarantees that the prime factor counts for 2 and 3 in

a^2b

meet or exceed the required thresholds, yielding a definitive 'Yes'.

Key Concept

Divisibility analysis using prime factor exponent inequalities in Data Sufficiency.

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