IIT JEE , AIEEE Forum - Mathematics , Algebra - (1) If a, b, c are natural no. than find out the probability of a2+b2+c2 is divisible by 7

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(1) If a, b, c are natural no. than find out the probability of a2+b2+c2 is divisible by 7.

(2) Find out the minimum non negative value of a, b, c if the polynomials x⁴+ax³+bx²+cx+d has real root.

(3) Find out the maximum value of (x-a)(x-b)(x-c)(ax+by+cz).If it is given that all a,b,c,(x-a),(x-b),(x-c) are positive.

(4) If f(x) = (a/x) + (b/x+b)-(c/x+c)-(d/x+d), and f(x) is divisible by x2, than find out the value of (1/a) + (1/b)-(1/c)-(1/d).

(5) Find all +ve integers a, b such that each of the equations x²-ax+b=0 and x²-bx+a=0 has distinct +ve integrals root.

hsbhatt

for the 1st one:

A number when divided by 7 leaves remainder r

(0,1,2,3,4,5,6).

So for a triad of numbers, the traid of remainders has 7*7*7 = 343 possibilties.

Now, the squares of numbers leave remainders 0,1,2 or 4 as remainders.

Hence if a²+b²+c² is div by 7 then either a,b and c are div by 7 or the remainders are permuations of (1,2,4).

The first case has one member in the set of triads of remainders.

Now, lets consider a²

1; b²

2 and c²

4 (mod 7)

You can check that number of favourable remainder triads is 8 for this case. and there are 6 such permutations of (1,2,4). Hence our count of favourable triads is 1+6*8 = 49.

Hence, the required prob = 49/343 = 1/7.

sboosy

A number when divided by seven can leave as remainder either one of

0,1,2,3,4,5,6

if u want the answer to b perfectly divisible by 7 it implies remainder=0

it is just a question of choosing 0 from the set 0,1,2,3,4,5,6

the probability of which is 1/7

hsbhatt

sboosy (wish I knew your actual name),

I agree that for a number in exclusion, any of the 7 remainders is equally likely. But, we cannot say that in this case each of the remainders is equally likely.

As a simple example what is the prob of a² leaving a remainder 3 on division by 7. Your answer is again 1/7. But no square leaves a remainder of 3 when divided by 7.

Again, what is the prob of a² leaving a remainder 1 on div by 7. This happens in two cases, when a

1or a

6 (mod 7). Hence the prob = 2/7.

sboosy

bhatt ji

i understand what u say.

i wrote that as the answer because when we add squares of numbers,

it could end with any of the numbers.

but when u consider only one number square

the answer is not same bcos squares of numbers do not end

with all possible numbers.

for example a number ending with 2 or 3 or 7 or 8 can never be a square.

anyway i understand ur point completely

hsbhatt

Sorry if I am bugging you, but I thought I should clarify this a bit

To take the point further, find the prob of the remainder being 1.

The favourable cases are permutations of (0,0,1) ; (0,4,4) and (2,2,4).

Case 1 occurs 3*2 = 6 times
Case 2 occurs 3*2*2 = 12 times
Case 3 occurs 3*2*2*2 = 24 times

Total = 42.

The prob = 42/343.

So, the remainders though being (0,1,..,6) they are all not equally likely

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