Messages posted by vorseraider1@gmail.com

it is 2(mod 3).

@tarun - google pe search karo bhai. there are methods to do that. simple methods.

A = (0,1,2,3,4,5,6) mod 7.

B = (0,1,2,3,4,5,6) mod 7.

so total ways to select A and B = 7*7 = 49.

favourable cases are (0,0), (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).

agey khud karlo.

i think this must be the way..

ha ha.

no need for wolfram alpha.

(1-9m)(1-m) = 385.

where d^2 = m.

from here we get, d^2 = 64/9 and -6.

so we get d = +8/3, -8/3, +iota*sqrt{6} and -iota*sqrt{6}.

wolfram alpha is the last resort...

sorry.

typing mistake.

w = complex cube root of unity.

i assume you mean w = complex root of unity.

w^3 = 1.

w + w^2 + 1 = 0

(a+b)(aw+bw^2)(aw^2+bw)

= (a+b)(a^2*w^3 + b^2*w^3 + ab*w^4 + abw^2)..

= (a+b)(a^2 + b^2 - ab)..

= a^3 + b^3.

caus w^4 = w..

the answer can not be 10!!!!!

N is 5(mod 9).

N = 9k + 5.

write 5 as 3 + 2.

N = 3(3k + 1) + 2.

so, N leaves the remainder 2 when divided by 3...

actually d = + 8/3, -8/3, +i*sqrt(6), -i*sqrt(6). where i = iota.

the r+1 th term in the expansion is C(10,r) * 5^(r/5) * 2^(5-r/2).

so, 2 divides r, 5 divides r, 10 divides r. so r = 0,10.

there are two such terms.

i had ot read the word with replacements.

so now i know what it means,

number of ways to pick up 7 coupons is 15^7. cause we have 15 choices on each time we pick.

there 9^7 ways to pick the coupons numbered 1 to 9.

we want at least one 9.

therefore number of ways = (total ways) - (ways where we don't have 9 and select from 1 to 8) = 9^7 - 8^7.

i didn't read the word - "replacement"!!!

shit...

number of ways to select 7 coupons = C(15,7).

we select 9.

the rest 6 to be selected from 1 - 8.

ways = C(8,6).

probabilty is C(8,6) / C(15,7) ???

galat ho sakta hai. probability nahi kari acche se..

see, 2009 is 2(mod 3).

and we raise it to an odd power,

2(mod 3) remains the same.

lemma -

2^(2n-1) is 2(mod 3) for all natural numbers , n.

4^n - 2 ,

(3+1)^n - 2.

(multiple of 3) - 1.

which is 2(mod 3). :)

so x = w^(power that is 2(mod3)).

x = w^2.

we want 1 + x + x^3 = 1 + w^2 + w^6 = 2 + w^2 = 1 + 1 + w + w^2 - w = 1 - w.

not sure. ;(

i think it should be 900 distinct N digit numbers. :)

w and w^2 are the roots of

1/(a + x^2) + 1/( b + x^2) + 1/(c + x^2) = 2x.

put x = 1, get RHS = 2.

:)

i think the question should be

let a,b,x,y satisfy ax+by=4 , ax^2 + by^2=-3 , ax^3+by^3=-3 then the value of (2x-1)(2y-1) is...

cause if it is by^3, it is too easy....

there is a set [1,-1,0] which keeps repeating.

the 108th tern is 0.

since 108 = divisible by 3.

like to kar de bhai!!

x = m^2.

x + 99 = m^2 + 99 = z^2.

z^2 - m^2 = 9*11.

(z-m)(z+m) = 3*3*11.

(z,m) = (1,10), (49,50), (15,18) and its permutations.

done.

i should have written.

y = 0, x = + , - root3.

x^2 + 7y^2 > 3. for all x , y = integers, and both x and y not simultaneously 0.

and if y = 0, x not equal to + 1 , - 1.

we can check these cases by hand.

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