Discussion

Explanation:

A = 1¹ × 2² × 3³ × 4⁴ × … 100¹⁰⁰

Product of one 2 and one 5 gives one trailing zero.

Here number of 2’s will definitely be more than number of 5’s, hence number of trailing zeros will be same as number of 5’s.

Powers of 5 will occur at 5⁵, 10¹⁰, 15¹⁵ … 100¹⁰⁰

Multiples of 5 
power at least 1 = 5 + 10 + 15 + … + 100 = 1050
additional power of 5 (i.e., power at least 2) = 25 + 50 + 75 + 100 = 250

Total number of 5s = 1050 + 250 = 1300

i.e., there will be 1300 trailing zeros in A.

Hence, 1300.

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