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(Q). Find the FNZD in 20!×30!×40!×50!×60!
Solution:- In this case the number of trailing zeros in the product will be the summation of number of trailing zeros in individual terms. & FNZD of the product will be the multiplication of the FNZD’s of all the factorial terms.
Hence

52053040506055555581102122540R0455555610000RRRR011031002022H = 4 7 9 12 142×4! 2×0!×1! 2×3! 2×2! 2×2!×2!16×24 8 2×6 6×2 4×2×2 4 2 2 6

∴ FNZD of product = 4×8×2×2×6
                                    = 8 Answer