211
factorial of nos after 9 hav 0 in their tens place
find the digit in tenth place of the sum1!+2!+3!+...........+49!
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4 Answers
1is this supposed to be solved like this by knowing this fact or is there some other way?
11!+2!+3!+4!=33, 5!=120, 6!=720, 7!=5040, 8! = 40320, 9! =326880 i.e the digit in tenth place of 1!+2!+.......+9! is 1 n! is divisible by 100 for all n ≥ 10 n!= 100kn for all n ≥10, where k10, k11, .........are positive integers.
i.e 10!+11! + .........+49!=100 (k10 +k11 +.......+k49)
i.e the digit in the tenth place of this sum is 0
i.e the digit in the tenth place of 1!+2!+......+49! is 1.