| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Jan 2008 14:30:11 IST
|
|
|
4 some natural number N,
the number of positive integral 'x' satisfying the eqn:::::
1! + 2! + 3! + ...........................+(x!) = N2 is:
|
this word is so small that it is a foolishness to hate anyone.
so, we love all. |
|
|
|
26 Jan 2008 14:54:56 IST
|
|
|
give the ans.
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
26 Jan 2008 16:26:25 IST
|
|
|
the answer is 2 ( for x= 1 and x=3).
1! = 1 = 12
1! + 2! = 3
1! + 2! + 3! = 9 = 32
1! + 2! + 3! + 4! = 33
1! + 2! + 3! + 4! + 5!= 153
From 5 onwards, all the factorials are divisible by 10.that means , the last digit in the sum of the given series remains same. but a square cannot have 3 as the last digit. so there are no more perfect squares in the series.
|
this reply: 4 points
(with 0
in 2 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
27 Jan 2008 15:12:32 IST
|
|
|
first of all the ans is :
2.....
and can any one plawese explain it to me.....
thnx in advance...........
|
this word is so small that it is a foolishness to hate anyone.
so, we love all. |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
28 Jan 2008 13:56:19 IST
|
|
|
nyone please??????????????
|
this word is so small that it is a foolishness to hate anyone.
so, we love all. |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
