First slide

Permutations

Question

The number of zeros at the end of 100! is

Moderate

A

36
B

18
C

24
D

None of these

Solution

In terms of prime factors 100! can be written as 2^a ⋅ 3^b ⋅5^c ⋅7^d …
Now, E₂ (100!)
$\begin{array}{l} = [\frac{100}{2}] + [\frac{100}{2^{2}}] + [\frac{100}{2^{3}}] + [\frac{100}{2^{4}}] + [\frac{100}{2^{5}}] + [\frac{100}{2^{6}}] \\ = 50 + 25 + 12 + 6 + 3 + 1 = 97 \end{array}$
and, $E_{5} (100!) = [\frac{100}{5}] + [\frac{100}{5^{2}}] = 20 + 4 = 24$
$\begin{matrix} 100! = 2^{97} \cdot 3^{b} \cdot 5^{24} \cdot 7^{d} \dots \\ = 2^{73} \cdot 3^{b} \cdot (2 \times 5)^{24} \cdot 7^{d} \dots \\ = 2^{73} \cdot 3^{b} \cdot (10)^{24} \cdot 7^{d} \dots \end{matrix}$
Hence, number of zeros at the end of 100! is 24.

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