IIT JEE , AIEEE Forum - Mathematics , Algebra

sandeepramesh

easy with mods :)

clearly not d

Ans is c clearly :D

sandeepramesh

note that both 115 and 96 are 1 mod 19 :D :D :D

sboosy

$[tex] \\ \mbox{Given} \ (115)^{96} - (96)^{115} \\ \\ = [(115)^{96} - (96)^{96}] - [(96)^{115} - (96)^{96}] \\ \\ a^n - b^n \ \mbox{is divisible by} \ a-b. \\ \\ \mbox{Hence} \ [(115)^{96} - (96)^{96}] \ \mbox{is divisible by} \ 115-96 = 19 \\ \\ \mbox{Now} \ [(96)^{115} - (96)^{96}] = ((96)^{96}) [ (96)^{19}-1] \\ \\ (96)^{19}-1 = (1+95)^{19} - 1 \ \mbox{On expanding ..the first term which is 1 ..cancels off } \\ \\ \mbox{leaving terms which are multiples of 95 and hence also 19} \\ \\ \mbox{Thus the first and second brackets both are divisible by 19} \\ \\ \mbox{Hence given is divisible by 19}$

vineetnegi

why not directly

19*5=95

19*6=114

(114+1)^96-(95+1)^115

using bionmial theorem for finding remainder

R(1)=1

R(2)=1

1-1=0

hence divisible

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