Question

If 100! = $2^a{3}^b{5}^c{7}^d...$, then

• A. $a=97$
• B. $b=\frac{1}{2}(a+1)$
• C. $c=\frac{1}{2}b$
• D. $d=\frac{1}{3}b$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $a=97$
C $c=\frac{1}{2}b$
D $d=\frac{1}{3}b$

i) no. of 2's in 100! are

$\left[\frac{100}{2}\right]+\left[\frac{100}{22}\right]+\left[\frac{100}{23}\right]+\left[\frac{100}{24}\right]+\left[\frac{100}{25}\right]+\left[\frac{100}{26}\right]$

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

$=97$ ie $a=97$

ii) no of 3's 100! are

$\left[\frac{100}{3}\right]+\left[\frac{100}{3^2}\right]+\left[\frac{100}{3^3}\right]+\left[\frac{100}{3^4}\right]$

$=33+11+3+1=48=6$

iii) no of 5's in 100 ! are

$\left[\frac{100}{5}\right]+\left[\frac{100}{5^2}\right]=20+4=24=c$

iv) no of 7's in 100 ! are

$\left[\frac{100}{7}\right]+\left[\frac{100}{7^2}\right]=14+2=16=d$

$c=\frac{b}{2}$ & $d=\frac{b}{3}$