Question

The largest natural number \(n\) such that \(3^n\) divides \(66!\) is:

Hint:

To find the highest power of a prime dividing \(n!\), use successive divisions by powers of the prime.

Answer 1

Correct Answer: 31

To determine the largest natural number \(n\) for which \(3^n\) divides \(66!\), we need to find the highest power of 3 that divides \(66!\). This can be calculated using the formula for finding powers of a prime \(p\) in \(n!\):

\[\sum_{k=1}^{\infty} \left\lfloor \frac{66}{3^k} \right\rfloor\]

For \(p=3\) and \(n=66\), calculate each term until the floor function results in zero:

• \(\left\lfloor \frac{66}{3^1} \right\rfloor = \left\lfloor 22 \right\rfloor = 22\)
• \(\left\lfloor \frac{66}{3^2} \right\rfloor = \left\lfloor \frac{66}{9} \right\rfloor = \left\lfloor 7.33 \right\rfloor = 7\)
• \(\left\lfloor \frac{66}{3^3} \right\rfloor = \left\lfloor \frac{66}{27} \right\rfloor = \left\lfloor 2.44 \right\rfloor = 2\)
• \(\left\lfloor \frac{66}{3^4} \right\rfloor = \left\lfloor \frac{66}{81} \right\rfloor = \left\lfloor 0.81 \right\rfloor = 0\)

Since subsequent terms are zero, we stop here. Summing these values gives:

\[22 + 7 + 2 = 31\]

Thus, the largest \(n\) such that \(3^n\) divides \(66!\) is 31.

Confirming against the provided range, \(31\) is indeed within the expected range of \([31, 31]\).