341what is the last digit for x>3?
Q. For some Natural No. N the no. of positive integral 'x' satisfying the equation
1! + 2! + 3! + ....... + (x!) = (N)2 is:
A) None B) one C) two D) infinite
I've found the values of x as 3 and 1. So total of 2 values is possible and the answer is also two.......BUT why only 2 solutions ????? Why is it not satisfied by any other no. ??? Please give the proof ...
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14 Answers
1I need help here !!!! Sending S.O.S for those who can receive the signals .... :-(
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.....
62yes subhomoy you are right :)
but you have to complete the proof by explaining the small ting that why no perfect square ends in 3?
49yeeeeeeeeee...............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???????
49but is there any other way of proving the above thing???????????
341I 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