The number of zeros at the end of 108! is

10

No worries! We‘ve got your back. Try BYJU‘S free classes today!

13

No worries! We‘ve got your back. Try BYJU‘S free classes today!

25

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

26

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $25$
Product of $5^{'} s$ & $2^{'} s$ constitute $0^{'} s$ at the end of a number $\Rightarrow$ No. of $0^{'} s$ in $108!$
$=$ exponent of $5$ in $108!$
(Note that exponent of $2$ will be more than exponent of $5$ in $108!$ )
$\Rightarrow [\frac{108}{5}] + [\frac{108}{5^{2}}] + [\frac{108}{5^{3}}] = 21 + 4 + 0 = 25$

Suggest Corrections

Similar questions

222^{111} \times 35^{53} + (7!)^{6!} \times (10!)^{5!} + 42^{42} \times 25^{25}

10

(5!)^{5!} + (10!)^{10!} + (50!)^{50!} + (100!)^{100!}

The number 1.2.3.4.5.6......25 has _______ number of zeroes in the end.

Join BYJU'S Learning Program