What is the highest power of 5 that divides of x = 100! = 100 x 99 x 98 x ...... x 3 x 2 x 1

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Solution

Calculating contributions of the different powers of 5 we
have $\frac{100}{5^{1}} = 20, \frac{100}{5^{2}} = 4$ hence
the total contributions to the power of 5 is 24 or the number 100! is
divisible by $5^{24}$

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Similar questions

5

x = 100!

f (x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \frac{x^{98}}{98}

, s h o w t h a t

f (x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \frac{x^{98}}{98} + \dots + x + 1,

f^{'} (1) = 100 f^{'} (0) .

x^{50}

(1 + x)^{100} + 2 x (1 + x)^{99} + 3 x^{2} (1 + x)^{98} + . . . . . . . . . + 101. x^{100}

x^{8}

(1 + x) (1 + x^{2}) (1 + x^{3}) (1 + x^{4}) (1 + x^{5}) . . .

. . . . (1 + x^{98}) (1 + x^{99}) (1 + x^{100})

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