106Q4. we need to find where the exponent of 5 = 26
till 100 it is 20+4 = 24..
now 105 has 25
and 110 has 26
so 110!, 111!, 112!, 113!, 114! have 26 zeroes
Q1 no of six digit nos which have sum of digits as odd integer
Q2 find Ak if \frac{m!}{x(x+1)(x+2)..(x+m)}=\sum_{i=1}^{m}{\frac{A_i}{x+i}}
Q3 largest two diigt prime number that divides^{200}C_{100}
Q4 no. of natural nos n for which n! ends with 26 zeroes
Q5 find no. f 3 digit nos .of the form xyz where x,z<y ,x≠0
Q6 what are ternery sequences ??
how many terenry sequences of length 9 are there which begin with 210 or end with 210
-
UP 0 DOWN 0 0 6
6 Answers
24Q7 x=(2n+1)(2n+3)...(4n-3)(4n-1);y=(\frac{1}{2})^n\frac{(4n)!.n!}{(2n)!(2n)!}
find value of x-y+2n
106
21Q3
200C100=200!100!.100!
in 200! der r exactly 2 nos which r multiple of 67,68,69.......100
whearas in 100! der is one multiple of 67,68......100...i.e the no. itself
des nos 67,68,69......100 gets cancellled from numerator & denominator in val of 200C100
in 200! der r exactly 3 nos which r multiply of 51,52..........66
wheras in 100! der is only one number multiple of 51,52....66(i.e no. itself)
so we hav 200C100=(61)3(No. not divisble by 61)(61x No. not divisble by 61)2
=61x integer
hence 61
1What is the sum of the digits
a > in 100000 --- ans. 1
b> in 100001 --- ans. 2
Take any arbitrary ,
c > in 123243 --- ans . 15
d > in 123244 --- ans . 16
So numbers which have odd sum of digits come right behind those who have even sum of digits .
What I want to say is , that the number of six digit numbers , that have odd sum of digits , is half the total no. of six digit nimbers.
So ans . --- 500000
1Given x = (2n+1)(2n+3) ........ (4n-3)(4n-1)
So x . (2n+2) . (2n+4) ..... 4n = (4n)! / (2n)!
Here we have n terms in L . H . S, starting from 2n+2 to ending with 4n . So by taking 2 from
each term , we get 2n [ (n+1)(n+2)(n+3) …. Upto 2n ] , which is (2n)! . 2n / n!
So x . (2n)! . 2n / n! = (4n)! / (2n)!
or x = 2-n . n! . 4nC2n -------------------- 1
Again , y = 4nC2n . n! . 2-n ------------- 2
So from 1 and 2 ,
y = x
Finally , x - y + 2n = 2n