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 Narsimha chary's messages in the community
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 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
show that 1111....(91 times) is a prime number
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p = 2008^2007 - 2008 and q =2008^2+2009 find the remainder when p divided by q
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solve for x, y; wherex^2 +4y^2-15x =10(3y -8) and xy = 6
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

This comes through extensive practice .

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

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.

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

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.

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

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
1994=(x+y)^2 + 3x +y; x,y are positive integers. Find y.
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

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.

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

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

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
three distinct numbers a,b,c form a G.P. Can a+b, b+c, c+a form an A.P.
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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
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
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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