Question

.
Find the number of trailing zeroes in 56!
(a) 13
(b) 11
(C) 12
(d) 6​

Answer 1

$\pink { Number \:of \: trailing \: zeroes \: in}\\\pink { a\:factorial \: (n!) }$

$\green { = Highest \: power \: of \:5 \: in \:n!}$

$Here, Given \: number \: 56!$

$i )\frac{[ 56]}{5} = \blue {11 }. It \: is \: greater \:than \:5$

$ii)\frac{[ 11]}{5} = \blue {2}. It \: is \: less \:than \:5$

$Now, we \: stop \: the \: process \:here .$

$\therefore \red{The \: trailing \:zeroes \: in \:56! }\\= 11 + 2 \\\green { = 13 }$

Therefore.,

$Option \: \pink { ( a ) } \:is \: correct.$

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Answer 2

Answer:13- Option 'a'

Step-by-step explanation: