Let N be the number of consecutive 0’s at the right end of the decimal representation of the product $1! 2! 3! 4! .....99! 100! .$ The remainder obtained when N is divided by 1000 is R, then which is are CORRECT?

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R is a two digit number

Number of positive integral divisors of R is 8

R is a three digit number

Number of positive integral divisors of R is 6

answer is A, D.

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Detailed Solution

Consecutive Zeros in Factorial Product

N = 1124

Remainder R = 124

Solution: To find trailing zeros, we count factors of 5 in each factorial (since 2's are more abundant).

• 96 factorials have at least one factor of 5 (initial count: 96)
• 91 have a factor of 10 (adds: 91)
• 86 have a factor of 15 (adds: 86)
• Continuing this pattern for all powers of 5
• Additional contributions: 76 + 51 + 26 + 1

Total: 970 + 76 + 51 + 26 + 1 = 1124 Therefore, N = 1124, and R = 1124 mod 1000 = 124.

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