Question

If \( a, b, c \) are all positive integers, with \( 4a>b \), then which of the following conditions is BOTH NECESSARY AND SUFFICIENT for the expression \[ \sqrt{(3)^a (21)^{3a-b} (49)^{2b+c}} \] to be a positive integer?

• A. \( a - b = c \)
• B. \( a - b + 2c \) is divisible by 3
• C. \( a, b, \) and \( c \) are divisible by 3
• D. \( a - b \) and \( c \) are divisible by 3
• E. None of the other conditions is both necessary and sufficient

Hint:

When working with powers, ensure that the exponents satisfy the conditions for the expression to be a perfect square, such as even exponents for all primes.

Answer 1

The Correct Option is B

Step 1: Express the given equation as a product of powers of prime factors.
We express each term in \( \sqrt{(3)^a (21)^{3a-b} (49)^{2b+c}} \) as powers of primes. - \( 3^a \) is already in prime factor form. - Since \( 21 = 3 \times 7 \), \( 21^{3a-b} = 3^{3a-b} \times 7^{3a-b} \). - Since \( 49 = 7^2 \), \( 49^{2b+c} = 7^{4b+2c} \). The expression becomes:\[\sqrt{3^a \times 3^{3a-b} \times 7^{3a-b} \times 7^{4b+2c}}\]Simplifying:\[\sqrt{3^{a + 3a - b} \times 7^{3a - b + 4b + 2c}} = \sqrt{3^{4a - b} \times 7^{3a + 3b + 2c}}\]
Step 2: Conditions for the expression to be a perfect square.
For the expression to be a perfect square, the exponents of all prime factors must be even. - The exponent \( 4a - b \) for prime 3 must be even. - The exponent \( 3a + 3b + 2c \) for prime 7 must be even.
Step 3: Analyze the conditions.
From \( 4a - b \) being even, it implies \( a - b \) must be even. From \( 3a + 3b + 2c \) being even, it implies \( a - b + 2c \) must be divisible by 3. This condition is both necessary and sufficient.
Final Answer: \[\boxed{\text{(B) } a - b + 2c \text{ is divisible by 3}}\]