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

To determine the necessary and sufficient condition for the expression: 

\(\sqrt{(3)^a (21)^{3a-b} (49)^{2b+c}}\)

to be a positive integer, let's break down the problem step-by-step:

1. Express the composite numbers \(21\) and \(49\) in terms of their prime factors:
   • \(21 = 3 \times 7\)
   • \(49 = 7^2\)
2. Substitute these in the given expression:
   • The expression becomes:
   • \(\sqrt{3^a (3 \times 7)^{3a-b} (7^2)^{2b+c}} = \sqrt{3^{a + (3a-b)} \times 7^{3a-b + 4b+2c}}\)
3. Simplify the exponents:
   • The exponent of \(3\) becomes: \(a + 3a - b = 4a - b\)
   • The exponent of \(7\) becomes: \(3a - b + 4b + 2c = 3a + 3b + 2c\)
4. Set the expression inside the square root to a perfect square:
   • For the whole expression to be a perfect square:
     • \(4a - b\) must be even (since it's exponent of 3).
     • \(3a + 3b + 2c\) should also be even.
5. Check divisibility by \(3\):
   • For simplicity, let's consider the modulo constraints:
     • Modulo 3 check on the condition \(a - b + 2c\) is needed:
     • If we have: \((4a - b) = 2k\) and \((3a + 3b + 2c) = 2m\).
     • Solving \((a - b + 2c) \equiv 0 \pmod{3}\) satisfies both conditions.

In conclusion, the condition "\(a - b + 2c\) is divisible by \(3\)" is both necessary and sufficient for the expression to be a positive integer. This is because it balances the parity requirements of exponents for both \(3\) and \(7\) to form a perfect square.

Answer 2

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