IIT JEE , AIEEE Forum - Mathematics , Algebra

sboosy

let a cubed = g
a square = h
given = pg+qh+ra+s
above div by 5 but s is not
which means that pg+qh+ra shud not be diiv by 5
taking a common we get
a(ph+qa+r) not div by 5
which means a not div by 5 and ph+qa+r not div by 5
now assume r = 5l ( multiple of 5)
which means ph+qa not div by 5
again taking a common
a(pa+q) not div by 5
a already not divisible so now pa+q not div by 5
assume q = 5i (multiple of 5)
which means pa not div by 5
so p not div by 5
now consider b cube = t
b square = e
given = st+re+qb+p (to show div by 5)
since p not divi by 5 as per above conclusion
TO SHOW implies st+re+qb not div by 5
taking b common
b(se+rb+q) not div by 5
b not div by 5 and se+rb+q not div by 5
since q div by 5
se+rb not div by 5
taking 5 common
b(sb+r) not div by 5
b not div and sb+r not div
since r div by 5
sb not div by 5
since s and b not div by 5
this is possible for some b not div by 5
hence proved

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