If 146! Is divisible by 5ⁿ, and then find the maximum value of n.

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Required answer, \\ = [\frac{146}{5}] + [\frac{146}{5^{2}}] + [\frac{146}{5^{3}}] \\ = 29 + 5 + 1 \\ = 35 \end{aligned}

$\eqalign{ & {\text{Required}}\,{\text{answer}}, \cr & = \left[ {\frac{{146}}{5}} \right] + \left[ {\frac{{146}}{{{5^2}}}} \right] + \left[ {\frac{{146}}{{{5^3}}}} \right] \cr & = 29 + 5 + 1 \cr & = 35 \cr}$
Note:
We have taken integral value only, not the fractional.
For example

\frac{146}{5}

$\frac{{146}}{5}$ = 29.2 but we have taken 29 and so on.

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Note:
We have taken integral value only, not the fractional.
For example

\frac{146}{5}

$\frac{{146}}{5}$ = 29.2 but we have taken 29 and so on.

", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Required answer, \\ = [\frac{146}{5}] + [\frac{146}{5^{2}}] + [\frac{146}{5^{3}}] \\ = 29 + 5 + 1 \\ = 35 \end{aligned}

$\eqalign{ & {\text{Required}}\,{\text{answer}}, \cr & = \left[ {\frac{{146}}{5}} \right] + \left[ {\frac{{146}}{{{5^2}}}} \right] + \left[ {\frac{{146}}{{{5^3}}}} \right] \cr & = 29 + 5 + 1 \cr & = 35 \cr}$
Note:
We have taken integral value only, not the fractional.
For example