official quant thread for cat08

[masoom]

Originally Posted by mahi101987

Lava prepares for a CAT exam by practising for 100 days . On any of these 100days she does not solve more than 20 questions. If on any day , she solves more than 12 questions then she solves at most 6 questions each on next two days . What is the maximum possible number of questions that she can solve over a period of 100 days

doing very freakily

... is it is 11*99+20...

[prakharc]

Originally Posted by implex

x^2-^2=k
(x+y)(x-y)=k
if k is od we are done
we can write it as k.1 and move
if k is even
if won't work
if it is not a power of 4 also
so
1000 nos can be divided as 4k,4k+1,4k+2+4k+3
all will work except for 4k+2
hence 750 solutions!!

Difference of squares is of the form 4k or 4k+1 .... can it be 4k+3 also??

[implex]

Originally Posted by prakharc

Difference of squares is of the form 4k or 4k+1 .... can it be 4k+3 also??

why it can't be ?
what about 4-1=3?

don't remember things man !!
difference of squares is not 4k or 4k+1
squares are 4k or 4k+1

[MaskedMenace]

Originally Posted by implex

x^2-^2=k
(x+y)(x-y)=k
if k is od we are done
we can write it as k.1 and move
if k is even
if won't work
if it is not a power of 4 also
so
1000 nos can be divided as 4k,4k+1,4k+2,4k+3
all will work except for 4k+2
hence 750 solutions!!

Implex bhaiya...nethor one..explain...bear me ...

[implex]

Originally Posted by MaskedMenace

Implex bhaiya...nethor one..explain...bear me ...

see (x+y)(x-y)=k.1
x-y=1 and x+y=k
x=(k+1)/2 and y=(k-1)/2

so if k is odd we r done !!
and similarly for others

Read my concept 1 on perfect squares on my quant blog
you will understand

[chintuz]

x^2-^2=k
(x+y)(x-y)=k
if k is od we are done
we can write it as k.1 and move
if k is even
if won't work
if it is not a power of 4 also
so
1000 nos can be divided as 4k,4k+1,4k+2,4k+3
all will work except for 4k+2
hence 750 solutions!!

Lets take example of 1, 2, 3, 4, 5, 6 and we see that difference of squares is .. 3, 5, 7, 8, 9, 12, 16, 21, 24, .. If we add more numbers we could end up with 4k, 4k+1, 4k+3 type of solutions with k=0 as start. Only numbers we will not get are 1 and 4. So it could be 748 solutions.

[MaskedMenace]

Originally Posted by implex

x^2-^2=k
(x+y)(x-y)=k
if k is od we are done
we can write it as k.1 and move
if k is even
if won't work
if it is not a power of 4 also
so
1000 nos can be divided as 4k,4k+1,4k+2+4k+3
all will work except for 4k+2
hence 750 solutions!!

sorry for troubling...explain tht bold part..n yeah..i ll read...
also any blog for DI ? If so please giv the link..

[implex]

Originally Posted by chintuz

Lets take example of 1, 2, 3, 4, 5, 6 and we see that difference of squares is .. 3, 5, 7, 8, 9, 12, 16, 21, 24, .. If we add more numbers we could end up with 4k, 4k+1, 4k+3 type of solutions with k=0 as start. Only numbers we will not get are 1 and 4. So it could be 748 solutions.

1 and 4 can be written as difference of perfect squares
1=1^2-0
and 4=2^2-0

0 is considered perfect square for such problems

[implex]

Originally Posted by MaskedMenace

sorry for troubling...explain tht bold part..n yeah..i ll read...
also any blog for DI ? If so please giv the link..

buddy
any number is of the form 4k,4k+1,4k+2 or 4k+3
for that matter out of every 4 consecutive nos, there will be one of each kind

[prakharc]

Originally Posted by implex

Problem from My blog
Let a,b,c be real numbers such that for all real numbers x, such that |x|<=1, we have |ax^2+bx+c|<=100. Determine the maximum value of |a|+|b|+|c|

@Implex :- There's a maximum "numerical" value and solution to this question naa(not some thing like cannot be determined etc..)??
No one was attempting/discussing this question so just asked to confirm ??.....

PS:- Don't post the solution....