Question

Rahul asked Shyam to find the smallest integer $N$ such that $N!>10^N$. Shyam says $N$ is between 10–15; Sohan says 16–20; Suresh says 21–25; Sonal says 26–31. Who is correct?

• A. Rahul
• B. Suresh
• C. Shyam
• D. None of these

Hint:

For inequalities of the form $N!$ vs.\ $a^N$, use $f(N+1)/f(N)=(N+1)/a$ to hop between $N$ values instead of recomputing factorials from scratch.

Answer 1

The Correct Option is B

Step 1: Compare via the ratio $f(N)=\dfrac{N!}{10^N}$.
Then $\displaystyle \frac{f(N+1)}{f(N)}=\frac{N+1}{10}$. Starting at $N=20$ (where values are easy to look up or estimate):
$20! \approx 2.4329\times10^{18}$, so $f(20)\approx \dfrac{2.4329\times10^{18}}{10^{20}}=0.024329<1$.
Step 2: Step upward using $\times\frac{N+1}{10}$.
$f(21)=f(20)\times\frac{21}{10}\approx 0.024329\times2.1\approx 0.0511$ (still $\lt 1$).
$f(22)=\times 2.2\approx 0.1124$; $f(23)=\times 2.3\approx 0.2584$;  $f(24)=\times 2.4\approx 0.6202$;  $f(25)=\times 2.5\approx 1.5505\ (>1)$.
Step 3: Threshold and conclusion.
$f(24)\lt 1$ but $f(25)\gt 1 \Rightarrow$ the smallest $N$ with $N!>10^N$ is $\boxed{25}$. This lies in 21–25, so Suresh is correct.
\[ \boxed{\text{Suresh}} \]