Messages posted by Narsimha chary

show that 1111....(91 times) is a prime number

p = 2008^2007 - 2008 and q =2008^2+2009 find the remainder when p divided by q

solve for x, y; wherex^2 +4y^2-15x =10(3y -8) and xy = 6

This comes through extensive practice .

2^192(256-31) + 2^n

2^192*225 +2^n ..........(1) , now let (1) be a perfect square and it is equal to m^2

m^2 - (2^96*15)^2 =2^n ; this is split into 2^n = (m + 2^96*15) * (m - 2^96 * 15) ................(2)

let n = 2A + B ; and substitute in (2)

gives 2^(A+ B) = m + 2^96*15 ..........(3) and 2^A = m - 2^96 *15 .......( 4)

solving equations 3 and 4 ; we have A = 97 and B = 4

therefore n = 2*97 +4 = 198

n = 198 gives the expression a perfect square.

x^100 = q(x) . (x-2)(x-1) + (Ax +B) ......................................(1) { remainder is 1 degree less than divisor.}

let x=2

substituting in (1)

we have

2^100 =2A +B .........(2)

let x = 1

on substitution in (1)

1 = A + B .........(3)then solving equations (2 ) and (3)

we have A = 2^100 -1 and B = 2 - 2^100

then remainder is (2^100 -1) x +2 - 2^100.

k.

it comes through quadratic formula

1994=(x+y)^2 + 3x +y; x,y are positive integers. Find y.

Use long division method and find the remainder. Later the divisor being a factor of dividend, remainder must be zero.

(By factor theorem)

but in this case it is 1+5k.

therefore 1+5k=0

5k = -1

k = -1/5.

okay.

the p,q,r terms of a GP and an HP are equal and are a,b,c respectively, prove that

a(b-c)log a + b(c-a)log b+c(a-b) log c=0

three distinct numbers a,b,c form a G.P. Can a+b, b+c, c+a form an A.P.

if a,b,c,d are in gp then show that (a^2-b^2) ,(b^2-b^2) ,(c^2-d^2) is in gp

if a,b,c,d are in gp then show that (a^2-b^2) ,(b^2-b^2) ,(c^2-d^2) is in gp

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