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a 0 1 2 3 4 5 6 7 8 9
a^5 0 1 2 3 4 5 6 7 8 9
so what i thaught was it should be true for greater no.s than 9 bcos the ending digit is same so the probablity of that m being a is more i tried some examples it was working fine.......tell me are there any violations to this condition.
v have to chk abt divisibility by 2,3,5
let me cal x as a^5 y as b^5 and z as c^5
div by 2:
if a,b,c are even x,y,z are even, so in this case div by 2 is satisied
if 1 is even 2 are odd,same will apply for x,y,z hence in this case also div by 2 is satisfied
div by 3:
a,b,c can be 0mod3 or 1mod3 or (-1)mod3 ie v select a,b,c from this such tht a+b+c is 0mod3
now if a is 0/1/(-1)mod3 x will also be 0/1/(-1)mod3 respectively..same can be said for b,c
wat i mean is wat ever remeinder a leaves on deviding by 3 same will be left by x..
now v have selected a,b,c such tht sum of the reminders when devided by 3,is a multiple of 3
now v c tht x,y,z also leave the same reminders..hence x+y+z will also leave reminder div by 3(ie x+y+z will be divisible by 3)
div by 5:
same concept as div by 3
a,b,c can be 0/1/(-1)/2/(-2) mod5 out of which the requisite a,b,c are selected..
now if a is 0/1/(-1)/2/(-2) mod5 then x will also be the same 0/1/(-1)/2/(-2) mod5..
same applies for b,c
so if v have a,b,c such thta+b+c has reminder as a multiple of 5 then x+y+z wud also have the same reminders summed up giving a reminder which is a multiple of 5...hence x+y+z wud be a multiple of 5...
so v see tht when a+b+c is multiple of 2,3,5 x+y+z will also be so..
hence if a+b+c=0mod30,x+y+z=0mod30
Gokul has indicated the right direction
The idea is to prove that n5-n is always divisible by 30
From Fermat's Little Theorem n5-n is always divisible by 5.
Again n5-n = n(n-1)(n+1)(n2+1)
n(n-1)(n+1) is divisible by 6 is easily proved (product of 3 consecutive numbers is divisble by 3! = 6)
This means n5-n is divisible by 30
Hence a5+b5+c5 -(a+b+c) is divisible by 30.
Hence if a+b+c is divisible by 30 so is a5+b5+c5
c^5 is divisible by 2, 3 and 5.
Look at possible values of a, b and c mod 2, 3 and 5. For example, mod 2
the possibilities for a, b and c respectively are:
0, 0, 0
1, 1, 1
0, 1, 1
1, 0, 0
(The order of a, b, c does not matter). You know that (a + b + c) mod 2 =
0, so the cases
1, 1, 1 and 1, 0, 0 can be eliminated. Plug the remaining cases into a^5 +
b^5 + c^5 and reach a conclusion.
Use the same method for mod 3 and mod 5.
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here we take
a+b+c=30k
(a+b+c)^5=30^5k^5
=a 5+b+c+5{a4(b+c)+a(b+c)4+bc4+cb4}+60{a3(b+c)2...........c2b3}...............(i)
here d terms wid 60 outside bracket r already div by 30..
for d term 5{a4(b+c)+a(b+c)4+bc4+cb4}
=5{a(b+c)(a3+(b+c)3)+bc(b3+c3)
=5(a(b+c)(a^3+(30k-a^)3+bc(b^3+c^3)
=in above expression all is div by 30 except, 5bc(b^3+c^3)
aagey................................pata nai............hehehe
if above xpression is div by 30............den its proved..............