IIT JEE , AIEEE Forum - Mathematics , Algebra

nitneh

it is 99/398

rebel

nopes dudes !!! tell me ur approach may be i can correct u !

aysh

The answer is 99/398

This iz how...

The expression 2^m+2^n+1 is divisible by 3 only when m and n are even.
Thus, reqd. no. of values of m & n=100x99
Also, total no. of possible values for m & n=200x199
Hence,
reqd. probability=(100x99)/(200x199)=99/398

Pl. rate ma hardwork people...

lord_maxima

I dont know if i am right or not,
so anyways

i found out(just by combining some numbers) that whenever m=n and m is a multiple of 2 say 2,4,6,8,... the no is always divisible by 3.

Say,m=n=2,
X=2^m+2^n+1
hence,X=9

Say,m=n=4,
X=33

But,m=n=3,
X=17
So,this means that no. of favourable outcomes will be >=100(2-2,4-4,6-6,....200-200)

Also,i figured out that pairing a multiple of 2(say 2 and 4),the number is divisible by 3 also.

Say,m=2,n=6,
X=69 => divisible by 3.

Anyways,just wanted to share with u all what i figured out but still cant get any answer.

lord_maxima

Hey,ignore my above attempt,
I want to ask that can m and n be equal like i used in my attempt???

rebel

sorry all wrong !

nitneh

if 99/398 i s wrong then tell me the right answer than i explain u the method

ramuramu

I think nitneh is right and if he is wrong tell him answer

harshiiit

yes for sure ans is 99/398

harshiiit

2power m =3-1power m
m and n are therefore even
p=100c2/200c2

ramyadiamond

Ok, i posted a wrong answer, bcoz i made a calculation mistake, but I've done it lke this:

The no.s which are divisible from 3 are formed when u take the same power of 2, i.e. same values of m and n, and they should be mulpitlpes of 2 only, like

2²+2²+1=9

2⁴+2⁴+1=33

2⁶+2⁶+1=129

Hence we see a pattern, in which if we choose both m and n as the same value, which are multiples of 2 only, in this way, we can choose 100 values. So, i think we've all done by the same method.

So, plz do tell us whats the right method to do it?

priyesh

for the no. to be divisible by 3

both m & n should be even (they may be same or may not be same)

therefore no. of ways of choosing m = 100 , n = 100

total ways 100 * 100

total ways of choosing without condition = 200 * 200 =

probability = (100 * 100)/(200*200) = 1/4

This soln. is valid only when selections are not neccesarily distinct but in this case since the two elements are distinct therefore answer will be

(¹⁰⁰C₂/ ²⁰⁰C₂) and not 1/4

dinesh_ddt

is it 1
boz 2^m+2^n is always div by 2 so it should be a sure event is it rite?

ramyadiamond

hey priyesh, i think it is 100*100/²⁰⁰C₂

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