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 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]

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...

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

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

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

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.

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

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.

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

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

shit...

 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]

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..

 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]

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. ;(

 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]

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

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

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.

:)

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

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....

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

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

the 108th tern is 0.

since 108 = divisible by 3.

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

like to kar de bhai!!

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

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.

 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]

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.

 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]

the question should be that these don't have integral solutions...

last one has a solution, y = 0. x = root3.

 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]

800000007 is 7(mod 8).

x^2, y^2 and z^2 can be 0,1,4(mod 8).

we can never get 7(mod 8) from them. :)

10511 is 3(mod 4).

a^2 and b^2 can be 0,1(mod 4).

again we can't get that..

 This Post 5 points    (1    in 1 votes )   [?]

hmmm.

motivational.

but i have my exam next year!!!
aur waise bhi top to karunga nahi...

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

we get

x + y + z + x^2(1/y + 1/z) + y^2(1/z + 1/x) + z^2 (1/x + 1/y) ...(1)

also, 1/x + 1/y + 1/z = 1/21.

in (1) we add and subtract x^2/x + y^2/y + z^2/z.

we get, (x^2+y^2+z^2)(1/x + 1/y + 1/z) = 1/21 * [5^2 + 2] = 27/21.

 Discussion Forums -> This Post 10 points    (2    in 2 votes )   [?]

@kareena.

are you really joking??

is this even a question??

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

no simple method to do it..

i don't understand a word of it..

:)

copied it from wolfram alpha so that people don't waste their time on this question.

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