What is the largest integer $n$ such that $11^n$ divides $1,000,000!$ ?

Answer 1

To determine the largest integer \( n \) such that \( 11^n \) divides \( 1,000,000! \), we need to find the highest power \( n \) of 11 that divides \( 1,000,000! \). This can be achieved by using the de Polignac's formula (Legendre's Formula), which is used to find the highest power of a prime \( p \) that divides \( n! \).

The formula is given by:
\[
\sum_{k=1}^{\infty} \left\lfloor \frac{N}{p^k} \right\rfloor
\]
where \( N = 1,000,000 \) and \( p = 11 \).

To calculate this, we proceed by evaluating each term in the sum until the terms become zero.

First, we calculate \( \left\lfloor \frac{1,000,000}{11} \right\rfloor \):
\[
\frac{1,000,000}{11} \approx 90909.0909
\]
Thus,
\[
\left\lfloor \frac{1,000,000}{11} \right\rfloor = 90909
\]

Next, we calculate \( \left\lfloor \frac{1,000,000}{11^2} \right\rfloor \):
\[
\frac{1,000,000}{121} \approx 8264.4628
\]
Thus,
\[
\left\lfloor \frac{1,000,000}{121} \right\rfloor = 8264
\]

Next, we calculate \( \left\lfloor \frac{1,000,000}{11^3} \right\rfloor \):
\[
\frac{1,000,000}{1331} \approx 751.6919
\]
Thus,
\[
\left\lfloor \frac{1,000,000}{1331} \right\rfloor = 751
\]

Next, we calculate \( \left\lfloor \frac{1,000,000}{11^4} \right\rfloor \):
\[
\frac{1,000,000}{14641} \approx 68.3152
\]
Thus,
\[
\left\lfloor \frac{1,000,000}{14641} \right\rfloor = 68
\]

Next, we calculate \( \left\lfloor \frac{1,000,000}{11^5} \right\rfloor \):
\[
\frac{1,000,000}{161051} \approx 6.2064
\]
Thus,
\[
\left\lfloor \frac{1,000,000}{161051} \right\rfloor = 6
\]

Next, we calculate \( \left\lfloor \frac{1,000,000}{11^6} \right\rfloor \):
\[
\frac{1,000,000}{1771561} \approx 0.5647
\]
Thus,
\[
\left\lfloor \frac{1,000,000}{1771561} \right\rfloor = 0
\]

Since higher powers of 11 will only result in terms of zero, we stop here and sum the terms calculated:

\[
\left\lfloor \frac{1,000,000}{11} \right\rfloor + \left\lfloor \frac{1,000,000}{121} \right\rfloor + \left\lfloor \frac{1,000,000}{1331} \right\rfloor + \left\lfloor \frac{1,000,000}{14641} \right\rfloor + \left\lfloor \frac{1,000,000}{161051} \right\rfloor = 90909 + 8264 + 751 + 68 + 6 = 100998
\]

Thus, the largest integer \( n \) such that \( 11^n \) divides \( 1,000,000! \) is:
\[
\boxed{100998}
\]