The number of zero at the end of 126! is

Solution:       $E_{2}$ = The exponent of 2 in $(126!)$

$E_{2}=[\frac{126}{2}]+[\frac{126}{2^{2}}]+[\frac{126}{2^{3}}]+[\frac{126}{2^4}]+[\frac{126}{2^{5}}]+[\frac{126}{2^{6}}]$

$E_{2}=63+31+15+7+3+1=120.$

$E_{5}$ =The exponent of 5 in $(126!)$

$E_{5}=[\frac{126}{5}]+[\frac{126}{5^{2}}]+[\frac{126}{5^{3}}]$

$E_{5}=25+5+1=31.$

$\therefore$       $E_{10}$ = The exponent of 10 in $(126!)$=$Min\{120,31\}=31.$

Most Viewed Questions

Preparation Products

Knockout JEE Main (Six Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 9999/- ₹ 8499/-
Knockout JEE Main (Nine Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 13999/- ₹ 12499/-
Knockout NEET (Six Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 9999/- ₹ 8499/-
Knockout NEET (Nine Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 13999/- ₹ 12499/-