IIT JEE , AIEEE Forum - Mathematics , Algebra

pramod6990

Let p,q,r,s be four integers such that s is not divisible by 5. If there is an integer a such that pa³+ qa²+ ra + s is divisible by 5, prove that there is an integer b such that sb³+ rb²+ qb + p is also divisible by 5.

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

hsbhatt

Every integer can be written in the form 5k, 5k

1, 5k

2

We are given that pa³+qa²+ra+s is divisible by 5 for some integer a. It is given that s is not divisible by 5 (otherwise, the problem would be trivial). Hence a

5k.

a = 5k

t, where t = 1 or 2

Case I:Now assume that a is of the form 5k+1

Then p+q+r+s is divisible by 5. In that case if we choose b = 5m+1

Then sb³+rb²+qb+p = multiple of 5 + (p+q+r+s) which is given to be a multiple of 5. So, for this choice of b, sb³+rb²+qb+p is divisible by 5.

Case II: a is of the form 5k-1. Hence -p+q-r+s is divisible by 5.

It is easy to see that again b = 5m-1 is the choice in this case

Case III: a is of the form 5k+2

Hence 8p+4q+2r+s is divisible by 5

Now choose b of the form 5k-2. The residue is -8s+4r-2q+p and our task is to prove that this expression is divisible by 5.

Consider 2(-8s+4r-2q+p ) + (8p+4q+2r+s) = -15s+10r+10p = 5n where n is an integer.

Also 8p+4q+2r+s = 5l

Hence 2(-8s+4r-2q+p ) = 5(n-l)

Since 5 does not divide 2, it divides (-8s+4r-2q+p )

Hence if a = 5k+2, choose b = 5m-2

Case IV: a = 5k-2

From the above working we can see that b must be of the form 5m+2

Hence for every choice of a, such that pa³+qa²+ra+s is divisible by 5, we can find infinite integers such that sb³+rb²+qb+p is divisible by 5.

hsbhatt

Please correct me if I am wrong, but I have a feeling that such problems have very little in common with a typical JEE qn.

I mean, they are looking for prospective engineers not number theorists.

Concentrate more on Calculus, Determinants, Coordinate Geometry, Vectors and Complex Numbers. A lot of maths that you learn in your 1st two years at IIT have to do with Calculus and Complex Numbers. Physics too begins with lot of Vector stuff. Coordinate Geometry is of enormous use to engineers. And the teachers in the Math departments teach and work in such areas. So these are the topics to look out for.

pramod6990

Im very sorry sir.........

u are 1250% right.....no need to argue......this was an INMO question that was tryin., got bugged by it and made a poor decision of posting it on the site......but never the less thnks for solving....and i can say that....

tussi great ho.....(for taking the effort and wasting time to solve such a lengthy problem)

hsbhatt

Of course, there's no harm in trying to find out the solution to such problems. Its not a waste of time at all. Who knows where it could come in use.

Just take it as advice from a well-meaning senior.

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