ISI2018-MMA-3

Question

The number of trailing zeros in $100!$ is

• A. $21$
• B. $23$
• C. $24$
• D. $25$

Comments

$\left \lfloor \dfrac{100}{5} \right \rfloor = 20$

$\left \lfloor \dfrac{20}{5} \right \rfloor = 4$

$\left \lfloor \dfrac{4}{5} \right \rfloor = 0$

$\therefore 20+4+0 = 24$

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https://gateoverflow.in/216621/isi-mtech-cse

Answer 1

To create 1 Zero  in 100! we need 5 and 2  so we have to check highest power of 2 and 5 in 100! , so follow the procedure below:

Finding highest power of 2 in 100!

Integer solution  of 100/2  + 100/2^2 + 100/ 2^3 + 100/2^4 + 100/2^5 +100/2^6 = 97
So highest power of 2 in  100!  Is 97

Now highest power of 5 in 100!
Integer solution of 100/5 + 100/5^2 = 20+4 =24
So in 100!  There  are 97  2's  and 24  5's
So we have less number of 5's   so no of fives will decide number  of trailing zeros in 100!   So Ans  is option C  that is 24

Answer 2

Answer is (C)
Multiple of 5 +Multiple of 25=20+4=24

Answer 3

Answer will be C

Answer 4

Let $f(n)$ give the number of trailing zeros in the base ten representation of $n!$ .

Then $f(n) = \left \lfloor \dfrac{n}{5} \right \rfloor + \left \lfloor \dfrac{n}{5^2} \right \rfloor + \left \lfloor \dfrac{n}{5^3} \right \rfloor+............+\left \lfloor \dfrac{n}{5^k} \right \rfloor$

where, $\left \lfloor \dfrac{n}{5^k} \right \rfloor = 0$

$f(n) = \left \lfloor \dfrac{100}{5} \right \rfloor + \left \lfloor \dfrac{100}{5^2} \right \rfloor + \left \lfloor \dfrac{100}{5^3} \right \rfloor$

$f(n) = \left \lfloor \dfrac{100}{5} \right \rfloor + \left \lfloor \dfrac{100}{25} \right \rfloor + \left \lfloor \dfrac{100}{125} \right \rfloor$

$f(n) = 20 + 4 + 0$

$f(n) = 24$