1I need help here !!!! Sending S.O.S for those who can receive the signals .... :-(
341what is the last digit for x>3?
341
Hari Shankar
·2009-11-11 00:30:02
why wink ?:D.
I 1st thot there was some mistake :)
62lol.. no there was none.. i just winked because i first thought that you gave a wrong hint :D
49the ans is clearly 2......1 and 3....
also for x≥4 1!+2!+3!.....+X!≠PERFECT SQR...
it can be proved...
1!+2!+3!+4!=1+2+6+24=33
from 5! onwards, every no ends with 0......
thus, \Sigma x! ends with 3 and cannot be a perfect square.........
am i correct???i am sure i am.....
62
Lokesh Verma
·2009-11-11 02:09:07
yes subhomoy you are right :)
but you have to complete the proof by explaining the small ting that why no perfect square ends in 3?
49
Subhomoy Bakshi
·2009-11-11 02:15:59
yeeeeeeeeee...............got it......
if i do like this is it ok?????
1*1=1
2*2=4
3*3=9
4*4=16
5*5=25
6*6=36
7*7=49
8*8=64
9*9=81
0*0=0
thus any number ending with 0 to 9 cannot hav sqare value ending with 3
and as every no mst end with 0 to 9 so no sqr ends with 3...
is this explanation ok???????
49
Subhomoy Bakshi
·2009-11-11 02:42:25
yyyyeeeeeeessssssssssssss.................
49
Subhomoy Bakshi
·2009-12-09 05:37:15
but is there any other way of proving the above thing???????????
341
Hari Shankar
·2010-03-05 22:39:27
I guess you are looking to prove that no square can have last digit 3.
i.e. there is no square of the form 10k+3. One way is of course to start with saying that every integer is of the form 10k±1, ±2,±3,±4 and look at the squares. But that is almost the same as what has been done in Post #9.
The shorter way is to use the Euler Totient Theorem:(read it up. Its a generalization of Fermat's Little Theorem).
We have \phi(10) = 4
Now we want to prove that we cannot have x^2 \equiv 3 \pmod {10}
Now , if x is even then x2-3 is an odd number and its not divisible by 10.
If x is a multiple of 5, evidently x2-3 is not divisible by 10.
So let us assume that x is prime to 10 and \phi(10) = 4
Then we have from Euler Totient Theoremx^4 \equiv 1 \pmod {10}
This implies 3^2 \equiv 1 \pmod {10} which contradicts 3^2 \equiv -1 \pmod {10}
So no perfect square has last digit 3
