Find
the common roots of the equation x2000+x2002+1
= 0 and x3+2x2+2x+1 = 0.
#IIT
#CMI #ISI
x3+2x2+2x+1
= 0
=>
x3+1+2x(x+1) = 0
=>
(x+1)(x2-x+1)+2x(x+1) = 0
=>
(x+1)(x2-x+1+2x) = 0
=>
(x+1)(x2+x+1) =0
=>
x = -1,w,w2 where 'w' is the cube root of unity
Now
putting these values in the expression x2000+x2002+1
we see that w and w2 only reduce it to zero. So
the common roots are w,w2
Find real x for which 1/[x] + 1/[2x] = {x} + 1/3, where [x] = greatest integer less than equal to x and {x} = x – {x}.
Let
x = z + f, where z is the integer part and 0 <= f < 1
Now
we have two cases (a) 0 <= f < 1/2 (b) 1/2 <= f < 1
Also
note that the R.H.S of the given equation i.e., {x} + 1/3 > 0 so
x has to be positive.
Case
(a) 0 <= f < 1/2, [x] = z and 2x = 2z + 2f , [2x] = 2z
since 0 <= 2f < 1
and
{x} = x – [x] = z + f – z = f.
This
reduces the equation to
1/z
+ 1/2z = f + 1/3 ........... (1)
=>
1/z + 1/2z >= 1/3 (Since f >= 0 )
=>
3/2z >= 1/3
=>
9 >= 2z
=>
2z – 9 <= 0
=>
z = 1,2,3,4
for
z = 1, {x} = f = 1 which is invalid in this case
for
z = 2, using equation (1) {x} = f = 5/12 < 1/2
for
z = 3, using equation (1) {x} = f = 1/6 < 1/2
for
z = 4, using equation (1) {x} = f = 1/24 < 1/2
so
possible values of x in this case is 2+5/12, 3+1/6 and 4+1/24
i.e.,
x = 29/12, 19/6 and 97/24
Case(b)
Following the steps of case (a)
show that in this case no solution exists.
Show
that abc(a³-b³)(b³-c³)(c³-a³) is always divisible by 7 where
a, b and c are non-negative integers.
Result:
Cube of any integer leaves remainder 0, 1 or 6 on dividing by 7.
if
any one of a,b or c is divisible by 7, then abc(a³-b³)(b³-c³)(c³-a³)
is divisible by 7. So we assume
that none of the numbers a,b or c is divisible by 7.
Now
using the result
stated above a³, b³,c³
leaves remainder 1 or 6 (due to the assumption) otherwise if 7
| a³ or
7 | b³
or 7 | c³
=> 7 | a or 7 | b
or 7 | c ( since 7 is prime) A contradiction to the assumption.
Now
by Pigeon Hole Principle two of the numbers a³, b³,c³ have the
same remainder. Hence their difference is divisible by 7, thus
7
| abc(a³-b³)(b³-c³)(c³-a³) in any case.
Find
the remainder when 2 is
divided by 1990
Result:
If p is prime & p do not divide a then ap
is congruent a (modulo p)
(Fermat's Theorem)
Note
that 1990 = 199 X 10 and 199 is a prime.
Using
Fermats theorem we have 2199
is congruent 2
(modulo 199)
=>
(2199
)10
is congruent 210
(modulo 199)
=>
2 is
congruent 1024 (modulo 199)
i.e
199 | 2 - 1024
Now
2
and
1024 has the
same last digit, which is 4 ( try to find it for 2
). Therefore
10
| 2
- 1024
Now
g.c.d (199, 10) = 1 therefore 1990 | 2
- 1024
So
the remainder is 1024
