The digit in tenth place of 1 ! + 2 ! + 3 + …… + 49 ! =

1 ! + 2 ! + 3 ! + 4 ! = 33 Also 5 ! = 120 , 6 ! = 720 , 7 ! = 5040 , 8 ! = 40320 and 9 ! = 326880 The digit in tenth place 1 ! + 2 ! + 3 ! + … + 9 ! = 1 Also note that n ! is divisible by 100 for n ≥ 10 so that the digit in tenth place 10 ! + 11 ! + … .. + 49 ! is zero. There fore The digit in tenth place of 1 ! + 2 ! + 3 ! + … .. + 49 ! is 1 .

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MathematicsProbabilityMiscellaneous/MixedMedium2 minQB

MathematicsMediuminteger

The digit in tenth place of $1! + 2! + 3 + \dots\dots + 49! =$

Answer:

Solution:

$1! + 2! + 3! + 4! = 33$ Also $5! = 120, 6! = 720, 7! = 5040, 8! = 40320$ and $9! = 326880$

The digit in tenth place $1! + 2! + 3! + \dots + 9! = 1$

Also note that $^{n!}$ is divisible by 100 for $n \geq 10$

so that the digit in tenth place $10! + 11! + \dots .. + 49!$ is zero.

There fore The digit in tenth place of $1! + 2! + 3! + \dots .. + 49!$ is 1 .

Stream:JEESubject:MathematicsTopic:ProbabilitySubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: QB

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