Number System

[tyro.novice]

Originally Posted by LearnerToro

Can u explain the first one in more detail.

Thank u

x^2-3x+2= (x-1) ( x-2)
x= 1,2
Put these values in x^100 - ax - b=0 Solve and u will get the values.

[toxic]

Quote:
Originally Posted by LearnerToro
Can u explain the first one in more detail.

Thank u

x^2-3x+2= (x-1) ( x-2)
x= 1,2
Put these values in x^100 - ax - b=0 Solve and u will get the values.

hey it's very simple.since remainderis ax+b,when x^100 is divided by x^2-3x+2 .that means x^100 -(ax+b) will be divisible by x^2-3x+2.
nd x^2-3x+2= (x-1)( x-2) so (x-1)( x-2) will also be factor of ax^100 -(ax+b) .so putting x=1,2 in
ax^100 -(ax+b)=0
we get the value of a & b.

[gujral17]

ans is -1 aor 16..solved using fremet theoram

[bleach it out]

this one is bugging me for week now...

consider the set Sn=(n,n+2,n+4,n+6,n+8 )
where n is natural number between 101 and 200(both inclusive)
how many of the 100 possible sets contain a multiple of 7...

[amitkrsingh]

Originally Posted by bleach it out

this one is bugging me for week now...

consider the set Sn=(n,n+2,n+4,n+6,n+8 )
where n is natural number between 101 and 200(both inclusive)
how many of the 100 possible sets contain a multiple of 7...

Let n=7m
Between 2 multiples of 7 there are 6 nos.

Now in order for Sn to contain multiple of 7 it can start with 7m+1,7m+3,7m+5 or 7m+6.

Smallest multiple of 7 between101 and 200 is 105.
and largest is 196.

Total 14 terms and 13 intervals(105-112,112-119, till 189-196) between them.

Each interval has 4 more sets which contain multiple of 7.

Hence total 14+13*4= 14+52 = 66 sets.

including sets begining 101,103,104,197, and 199 there are total 71 sets between 101 and 200 which contain multiple of 7.

[Flameboy007]

Originally Posted by bleach it out

this one is bugging me for week now...

consider the set Sn=(n,n+2,n+4,n+6,n+8 )
where n is natural number between 101 and 200(both inclusive)
how many of the 100 possible sets contain a multiple of 7...

B/w 101 and 200, mutiplez of 7 are:
105 = 7*15
...
189 = 7*27
196 = 7*28

Considering,
n as a multiple of 7, there are (27-15) = 13 sets (including 27 and 15)
n + 2 as a multiple of 7, there are (27-15) = 13 sets (including 27 and 15)
n +4 as a multiple of 7, there are (28-15) = 14 sets (including 28 and 15)
n +6 as a multiple of 7, there are (28-16) = 13 sets (including 28 and 16)
n +8 as a multiple of 7, there are (28-16) = 13 sets (including 28 and 16)

Hence, a total of 13*4 + 14 = 66 sets

[prakharc]

The remainder when 2^2+22^2+222^2+.......(222.......49twos)^2 is divided by 9 is :

(a)2
(b)5
(c)6
(d)7

I m getting answer as 4.......don't know where I am going wrong??

[MaskedMenace]

Originally Posted by prakharc

The remainder when 2^2+22^2+222^2+.......(222.......49twos)^2 is divided by 9 is :

(a)2
(b)5
(c)6
(d)7

I m getting answer as 4.......don't know where I am going wrong??

hii..

can u tel me how to proceed ?

[prakharc]

Originally Posted by MaskedMenace

hii..

can u tel me how to proceed ?

take 4 outside..you'll get series of the form 1+11^2+111^2+.....)
if you take remainde rof each term by 9...you'l get remainder of form...1*1,2*2,3*3,4*4,5*5......which forms repeating pattern after 8 or 9 terms

[shashank3012]

Originally Posted by prakharc

The remainder when 2^2+22^2+222^2+.......(222.......49twos)^2 is divided by 9 is :

(a)2
(b)5
(c)6
(d)7

I m getting answer as 4.......don't know where I am going wrong??

Is the answer 6?