New Problems
Hi if u can solve this problems thn plz come up with soln. & submit the way to approach....
Q1. 1000!=(100!)^k - P Where P is a prime number & both P,k are integers. Find the maximum value of k.
(a)105, (b)106, (c)104, (d)102,(e)None.
Q2.What is the sum of the sum of the digits of 55!.
(a)36, (b)25, (c)9, (d)5.
Q3.Find the remainder of 40!/83.
(a)45,(b)81,(c)65,(d)1.
n/a
Q2.What is the sum of the sum of the digits of 55!.
(a)36, (b)25, (c)9, (d)5.
ANS:
I think its 9
all the numbers ≥ 6 gives 9
i.e Sum of(Sum of(digits of n!)) = 9 where n ≥ 6
I think at the primitive stage of ths problm as u hv taken (p-1)....So at the last time u should take (p-2) instead of takin only -2....
Thankin you,
Dipanjan
n/a
Can any 1 solv the 1st problm wth explanation....l
n/a
10000!/100! let assume a small no. 100!/10! 1! to 10! will be divided once by denominator now from (11! to 20!) /10! will give in numerator 2*2 and some prime no similerly from 20! to 30! it will give 3 * 3 and some prime no, so upto 100! it will be devided by 10 times by 10!
now in numerator 1*1*2*2*3*3.............10*10 this will be devided by 2 times more by 10! so answer will be 12
so the original que answer is 102
That's correct and reason is (sum of digits (sum of digits(n!)))
of any number >= six is divisible by 9 as well.
Why it should not be 2 . . . .take it as an assignment