Posts and Comments

Sachin Kaushik
@Skaushi
14

@ayushbhalotia 49: it comes out to be 24. I can't make out of proportional sign for 0....

@ayushbhalotia 49: it comes out to be 24. I can't make out of proportional sign for 0....

Sachin Kaushik
@Skaushi
14

@[586289:abhimanyu03] Well, I chose the latter part ...quit onsite and started preparing...thinking about you...well, you know how it feels to get 99....so don't think too much...take the plunge...you have done it once , you can very well do it twice....technology...however niche...will eventuall...

@[586289:abhimanyu03] Well, I chose the latter part ...quit onsite and started preparing...thinking about you...well, you know how it feels to get 99....so don't think too much...take the plunge...you have done it once , you can very well do it twice....technology...however niche...will eventually fade out....so i ll suggest you to come back...

Sachin Kaushik
@Skaushi
14

> Q2
*scoring 60percent=losing 40 percent=100 marks
hence a+b+c+d=100
hence we have 103C3 solns..
but a,b,c cannot be >50
hence a'+b+c+d=49
we have 52C3 solns
hence 103C3-3*52C3=110551*
*
Q4...The possibilities are
case1..all similar
only 9 cases are possible
...

Q2scoring 60percent=losing 40 percent=100 marks

hence a+b+c+d=100

hence we have 103C3 solns..

but a,b,c cannot be >50

hence a'+b+c+d=49

we have 52C3 solns

hence 103C3-3*52C3=110551

Q4...The possibilities are

case1..all similar

only 9 cases are possible

case2 2 distinct digits present...there are sub cases to this

A>when none are 0

9C2(2*2*2*2-2)=504

B>when 0 is one of the digits

9C1(1*2*2*2-1)=63

hence total cases=504+72=576

Sam sir, can you explain a bit....a'+b+c+d=49 step...

Sachin Kaushik
@Skaushi
14

> Let abcd be the four digit number
Now, either first three digit will be consecutive or last three
Case-1: First three digit is consecutive
It is also divisible by 9
=> a+a+1+a+2+b=9k
=> 3a+b=9k-3
Now, maximum value of 3a+b=36
If 3a+b=6 (k=1)
=> (a,b)= (1,3)
If 3a+b=15 (k=2)...

Let abcd be the four digit number

Now, either first three digit will be consecutive or last three

Case-1: First three digit is consecutive

It is also divisible by 9

=> a+a+1+a+2+b=9k

=> 3a+b=9k-3

Now, maximum value of 3a+b=36

If 3a+b=6 (k=1)

=> (a,b)= (1,3)

If 3a+b=15 (k=2)

=> (a,b)= (4,3),(6,3)

If 3a+b=24 (k=3)

=> (a,b)= (7,3),(6,6)

Thus, total 5 cases

Case 2:- Last three digits are consecutive

=> a+3b= 9k-3

If a+3b=6

2 cases

If a+3b=15

2 cases

If a+3b=24

2 cases

Thus, total 11 cases

P.S:- Neetuji Pls check. I think i missed something

I think 14 cases will be there in total. Cases where b=0 (considering it is always the one's place digit) will be counted as 4 digit numbers as well.

Sachin Kaushik
@Skaushi
14

> 28 = 7*4 or 7*2*2
24 = 8*3
from 28 factors , it has reduced to 24 factors.
Thus, power of 2 in first case is 3, which reduces to 2 in second case.
And power of 3 increases from 6 to 7.
Thus, 2n^2 = 2^3 * 3^6
3n^2 = 2^2 * 3^7
thus, 6n^2 = 2^3 * 3^7
*
Factors = 4*8 ...

28 = 7*4 or 7*2*2

24 = 8*3

from 28 factors , it has reduced to 24 factors.

Thus, power of 2 in first case is 3, which reduces to 2 in second case.

And power of 3 increases from 6 to 7.

Thus, 2n^2 = 2^3 * 3^6

3n^2 = 2^2 * 3^7

thus, 6n^2 = 2^3 * 3^7

Factors = 4*8 = 32

Vikram , could you elaborate a bit.. I have not been able to understand the approach...

Sachin Kaushik
@Skaushi
14

> awesomeusername Says
>
> yup..how did u get it
1 approach could be start with 5 variables, a,b,c,d,e in incrreasing order and now make eqns, eg highest 2 d+e=121 a+b=111 and so on....

awesomeusername Saysyup..how did u get it

1 approach could be start with 5 variables, a,b,c,d,e in incrreasing order and now make eqns, eg highest 2 d+e=121 a+b=111 and so on....

Sachin Kaushik
@Skaushi
14

> Ques 1)
X+Y<=65
if X=0 Y<=65
4X+3Y=221
DIVIDE BY 4 AND TAKE ONLY REMAINDERS
Y/4=1 (value of Y such that remainders=1 always)
so Y can take 1,5,9,...,65
so total no. of solution are 17.
Rajan, could you explain a bit bout division by 4 here...why we are doing it ?

Ques 1)

X+Y<=65

if X=0 Y<=65

4X+3Y=221

DIVIDE BY 4 AND TAKE ONLY REMAINDERS

Y/4=1 (value of Y such that remainders=1 always)

so Y can take 1,5,9,...,65

so total no. of solution are 17.

Rajan, could you explain a bit bout division by 4 here...why we are doing it ?

Sachin Kaushik
@Skaushi
14

> My take:b
10C2*2!*2!*9!
=>18*10!
Please thoda explain karo bhai...What stands for what ?

My take:b

10C2*2!*2!*9!

=>18*10!

Please thoda explain karo bhai...What stands for what ?

Jatin_kaushik Says6^n:banghead:

I think it ll be 6*(N-1)

Sachin Kaushik
@Skaushi
14

> arumugadas Says
>
> A & B start at one end whose len 50 m long ...the race is 1000m ,,if A beats B and meets him 17 times during course and A speed is 5 m/s then..B?
Is 4.25 ms the answer ?

arumugadas SaysA & B start at one end whose len 50 m long ...the race is 1000m ,,if A beats B and meets him 17 times during course and A speed is 5 m/s then..B?

Is 4.25 ms the answer ?

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