30no one?
5 Answers
1we can solve it this like this........
510 - 310
=(52)5 - (35)2
=(11*2 +3)5 - (11*22 + 1)2
=(11q + 35) - (11p +12)
=11q +11p +242
=11(q+p+22)
1well from eulers theorem since 5 and 11 r coprime so 5^{\varphi (11)}\equiv 1mod11
\varphi (11)=10
so 5^10≡1mod11
similarly 3 and 11 r coprime so 3^10=1mod11
so 5^10-3^10=(1-1)mod11≡0mod11
1153≡4(mod 11)
(53)3≡43(mod 11)
510≡43X5(mod 11)
43X5≡1(mod 11)
Therefore,510≡1(mod 11)
35≡1(mod 11)
(35)2≡12(mod 11)
310≡1(mod 11)
510-310≡1-1(mod 11)
510-310≡0(mod 11)
OR
Since
510≡1(mod 11)
Let 510 be 11 k+1
310≡1(mod 11)
Let 310 be 11 m+1
Therefore
510 - 310 =11k+1 -11m-1
=11(k+m)
Hence it is divisible by 11
11There was just a small error which I have corrected:
11k+1-11m-1
=11(k-m)