if n greater than 100 and sum of 101 102 ------ n what is the the probability that the sum

@rakesh- when sum has a factor of (n-100) in it, that itself means it is completely divisible by (n-100).
So, don't you think it is completely divisible, independent of value of n, so probability is 1.
10 years agoHelpfull: Yes(17) No(9)
a=101,d=1;
Sn = n/2(2*101+n-1)
Sn= (n^2+201n)/2

let chk Sn divisible by (n-100)
(n^2+201n)/2(n-100)

depend n is even or odd but Denominator is even.
if n is even--- even/even
else n is odd ---- odd/even
so C)1/2 may be right ans....
10 years agoHelpfull: Yes(6) No(2)
@trisha 1/2*(n+101).....this must be a integer for sum to have
n-100 as factor so for 1/2*(n+101) being an integer n+101 must be a even no. so
probability is 1/2
10 years agoHelpfull: Yes(2) No(0)
sum=[n-101+1][101+n]/2
=(n-100)(101+n)/2
prob={(n-100)(101+n)/2}/n-100
so,ans is=D
10 years agoHelpfull: Yes(1) No(5)
@trisha .. probablity cant be 1
bcozz take example of 106 . dis doesn't agree the term
10 years agoHelpfull: Yes(0) No(0)

if n greater than 100 and sum of 101+102+------+n, what is the the probability that the sum will be divisible by (n-100)
a) 3/4
b) 1/3
c) 1/2
d) None of above