Question

If a and b are positive integers satisfying \(4a + 3b = 17\) and \( \frac{(5ab)!}{3^p} \) is an integer, then the maximum value of p is

• A. 13
• B. 14
• C. 15
• D. 16
• E. 17

Hint:

For linear Diophantine equations like \(ax + by = c\) with small coefficients, systematically testing integer values for one variable is the fastest way to find solutions, especially when the variables are constrained to be positive.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem has two parts. First, we need to find the integer solutions for a linear Diophantine equation with constraints. Second, we need to find the highest power of a prime number (3 in this case) that divides a factorial number, which can be found using Legendre's formula.
Step 2: Key Formula or Approach:
To find the highest power of a prime \(p\) in \(n!\), we use Legendre's Formula: \[ E_p(n!) = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \dots \] Step 3: Detailed Explanation:
Part 1: Find integer solutions for a and b.
We are given the equation \(4a + 3b = 17\), where \(a\) and \(b\) are positive integers.
We can test values for \(a\) starting from 1.
If \(a=1\): \(4(1) + 3b = 17 \implies 3b = 13 \). \(b\) is not an integer.
If \(a=2\): \(4(2) + 3b = 17 \implies 8 + 3b = 17 \implies 3b = 9 \implies b = 3\). This is a valid solution.
If \(a=3\): \(4(3) + 3b = 17 \implies 12 + 3b = 17 \implies 3b = 5 \). \(b\) is not an integer.
If \(a=4\): \(4(4) + 3b = 17 \implies 16 + 3b = 17 \implies 3b = 1 \). \(b\) is not an integer.
If \(a \geq 5\), \(4a \geq 20\), which would make \(3b\) negative. Since \(b\) must be positive, there are no more solutions.
So, the only pair of positive integers \((a, b)\) is \((2, 3)\).
Part 2: Find the maximum value of p.
The condition that \( \frac{(5ab)!}{3^p} \) is an integer means that \(p\) is the exponent of the highest power of 3 that divides \((5ab)!\).
First, calculate \(5ab\):
\[ 5ab = 5 \times 2 \times 3 = 30 \] So we need to find the highest power of 3 in \(30!\).
Using Legendre's formula with \(n=30\) and prime \(p=3\):
\[ E_3(30!) = \left\lfloor \frac{30}{3} \right\rfloor + \left\lfloor \frac{30}{3^2} \right\rfloor + \left\lfloor \frac{30}{3^3} \right\rfloor + \dots \] \[ E_3(30!) = \left\lfloor \frac{30}{3} \right\rfloor + \left\lfloor \frac{30}{9} \right\rfloor + \left\lfloor \frac{30}{27} \right\rfloor \] The next term \(\left\lfloor \frac{30}{81} \right\rfloor\) will be 0, so we can stop.
\[ E_3(30!) = 10 + 3 + 1 = 14 \] Step 4: Final Answer:
The maximum value of \(p\) is 14.