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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...
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actually d = + 8/3, -8/3, +i*sqrt(6), -i*sqrt(6). where i = iota.
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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.
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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.
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i didn't read the word - "replacement"!!! shit...
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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..
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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. ;(
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i think it should be 900 distinct N digit numbers. :)
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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. :)
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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....
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there is a set [1,-1,0] which keeps repeating. the 108th tern is 0. since 108 = divisible by 3.
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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.
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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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the question should be that these don't have integral solutions... last one has a solution, y = 0. x = root3.
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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..
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hmmm. motivational. but i have my exam next year!!!
aur waise bhi top to karunga nahi...
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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.
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@kareena. are you really joking?? is this even a question??
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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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Vriti Edustore
