The tens place digit of $1! + 2! +________+ 29!$ is:

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answer is C.

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Detailed Solution

Concept- We will first expand the first ten factorials of the series to get after 4! and the remaining factorials are 0, therefore we will add the unit digit of 1! 2! 3! 4! and here we will get a double-digit number from the addition of unit digits then, on keeping the tens place number from this double digit and adding in the number which comes from the addition of tens place digit from 4! to 9! and from 9! Onwards, the tens place value of the digit is 0, therefore adding the tens value of these numbers will not bring any change. Expanding first 10 factorials starting 1:

1! = 1 2! = 2 3! = 6 4! = 24 5! = 120 6! = 720 7! = 5040 8! = 40320 9! = 362880 10! = 3628800

On seeing, the first 4 factorials are ending with a non-zero-unit digit, therefore we are adding the unit digit of these numbers as,
Adding unit place digits

= 1 + 2 + 6 + 24

Adding unit place digits

= 13 … … i

Now, from 4! To 9!, we have tens place as non-zero, i.e.

120 + 720 + 5040 + 40320 + 362880

Therefore, adding the tens place digits of these numbers as,
Adding unit place digits

= 2 + 2 + 2 + 4 + 2 + 8

Adding unit place digits

= 20 … … i i

Now, from the equation (1), we will add the tens digit of the number to 13 in equation(ii), i.e.

13 + 20

as the question demands the tens digit of the factorial series. On adding 1 and 20 gives 21, so the tens digit is 1.
Hence, the correct answer is option 3.

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