The number of zeros at the end of $200!$ , is

49

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48

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50

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51

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Solution

The correct option is A $49$
Multiples of $2 = (2, 4, 6........198, 200)$
Multiples of $5 = (5, 10, 15, . . . . . . . .195, 200)$
Number of zeros
$\Rightarrow \frac{200}{5} = 40$
But for numbers like $25, 125$
$\Rightarrow 25 = 5 \times 5 \Rightarrow 125 = 5 \times 5 \times 5$
And for numbers like $4, 8, 16$
$\Rightarrow 4 = 2 \times 2 \Rightarrow 8 = 2 \times 2 \times 2$
Total number of zeros $= 40 + 8 + 1 = 49$
Therefore number of trailing zeros $= 49$
Hence the correct answer is $49$

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Similar questions

51^{49} + 51^{48} + 51^{47} + . . . + 51 + 1

Question 12

The successor of, the predecessor of -50 is

(a) -48

(b) -49

(d) -51

\frac{^{50} C_{0}}{1}, \frac{^{50} C_{2}}{3}, \frac{^{50} C_{4}}{5}, \dots, \frac{^{50} C_{50}}{51}

Weight (in kg)	48	49	50	51	52	53
Number of Students	4	10	20	11	3	2

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

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