The unit digit in 1 ! + 2 ! + 3 ! + …… + 49 ! is

1 ! + 2 ! + 3 ! + 4 ! = 33 5 ! = 120 , 6 ! = 720 , 7 ! = 5040 8 ! = 40320 , 9 ! = 326880 Thus the two digit of 1 ! + 2 ! + … . + 9 ! = 1 Also note that n ! is divisible by 100 for all n ≥ 10 . ∴ term digits of 10 ! + 11 ! + …… + 49 ! = 0 ∴ term digits of 1 ! + 2 ! + …… + 49 ! = 1 .

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MathematicsPermutation & CombinationFundamental PrincipleEasy2 minQB

MathematicsEasyinteger

The unit digit in $1! + 2! + 3! + \dots\dots + 49$ ! is

Answer:

Solution:

$1! + 2! + 3! + 4! = 33$

$5! = 120, 6! = 720, 7! = 5040$

$8! = 40320, 9! = 326880$

Thus the two digit of

$1! + 2! + \dots . + 9! = 1$

Also note that $n!$ is divisible by 100 for all $n \geq 10$ .

∴ term digits of $10! + 11! + \dots\dots + 49! = 0$

∴ term digits of $1! + 2! + \dots\dots + 49! = 1$ .

Stream:JEESubject:MathematicsTopic:Permutation & CombinationSubtopic:Fundamental Principle

⏱ 2mℹ️ Source: QB

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