Question

Let $N=\frac{100!}{{20}^m}.$ The smallest positive integer $m$ for which $N$ is not an integer is
​​​​​​​(correct answer + 1, wrong answer - 0.25)

• A. $5$
• B. $6$
• C. $24$
• D. $25$

Answer 1

The correct option is D $25$

${20}^m=2^{2m}\times 5^m$
We find the exponent of $2,5$ in $100!$

Exponent of $2,$
$m_1=\left[\frac{100}{2}\right]+\left[\frac{100}{2^2}\right]+\left[\frac{100}{2^3}\right]+\cdots$

$=50+25+12+6+3+1$
$=97$

Exponent of $5,$
$m_3=\left[\frac{100}{5}\right]+\left[\frac{100}{5^2}\right]+\left[\frac{100}{5^3}\right]+\cdots$

$=20+4$
$=24$

$N$ is not an integer if
$2m>97\text{or}m>24$
i.e., $m>48.5\text{or}m>24$
Therefore, the smallest positive integer $m$ for which $N$ is not an integer is $25.$