official quant thread for cat08

[naga25french]

Originally Posted by ahaprashant

Couple of small doubts, are you sure for ANY number (7 / 8 / 459 / 2463 etc. ), we have last two digits repeating every 20 powers!??? I knew it was true for some numbers, didn't realize this universality!

yeah for any number from 1 to n , the last two digits will have a cyclicity of 20... this will surely work for any number..

Originally Posted by ahaprashant

Secondly, by saying "dividing by 100", you mean divide the power with 100, then simplify right ? (which is nothing but making us eof the fact that last two digs repeat after 20 power) ? RIGHT ?

i guess u interpreted wrongly.. let me give u an eg.

find last two digit of 7^881

first calculate 881 mod 20 = 1

next step is to calculate 7^1 /100 = 07

this is what is meant by division by 100...

hope u get it now

[ahaprashant]

Originally Posted by Vishal2704

Is the 4th one 72 ??...already edited d post

Nope, try again.

[Vishal2704]

Originally Posted by ahaprashant

Nope, try again.

are yar fir galtiii

92 tha....agar ye bhi galat to jaldi se answer batao

[ahaprashant]

Originally Posted by naga25french

yeah for any number from 1 to n , the last two digits will have a cyclicity of 20... this will surely work for any number..




i guess u interpreted wrongly.. let me give u an eg.

find last two digit of 7^881

first calculate 881 mod 20 = 1

next step is to calculate 7^1 /100 = 07

this is what is meant by division by 100...

hope u get it now

Yep got it now, thanks a lot.
Btw a small correction, your fourth answer is wrong. Plz check again.

[ahaprashant]

Originally Posted by Vishal2704

are yar fir galtiii

92 tha....agar ye bhi galat to jaldi se answer batao

Correct

[ikruz]

Hi puys ,

I have a doubt in the following question -

Minimum of 3x+4y,subject to the condition (x^2)(y^3)=6 and x,y positive is ?

Pooja's answer is as follows

Originally Posted by poojadatta

3x/2=4y/3= k
x= 2k/3 and y= 3k/4
putting this in x^2*y^3=6 we get k=2
x=4/3 and y=3/2
putting this in 3x+4y we get 10

But i didnt understand why 3x/2=4y/3=k was taken at the first place?.

Can anyone please explain in a simpler way?.

[ahaprashant]

Originally Posted by ikruz

Hi puys ,

I have a doubt in the following question -

Minimum of 3x+4y,subject to the condition (x^2)(y^3)=6 and x,y positive is ?

Pooja's answer is as follows




But i didnt understand why 3x/2=4y/3=k was taken at the first place?.

Can anyone please explain in a simpler way?.

Hello,
It's an AM>GM problem wherein you've to decide which possible combinations to take to form AM & GM and finally apply the inequality,
as in this case

(3x/2 + 3x/2 + 4y/3 +4y/3 +4y/3)/ 5 > (3x/2 x 3x/2 x 4y/3 x4y/3 x 4y/3) ^1/5 giving (3x + 4y) / 5 > (16 (x^2)(y^3)) ^1/5
OR, (3x + 4y) > 5 (16 (x^2)(y^3))^1/5 = 5(96)^1/5 = 10 (3)1/5
Hope that helps!

Pagal Gadha

[ikruz]

The question is to find out maximum value of 2x^2 + 5y^2 + 8z^2 ,if 2x+3y+4z=100.

Originally Posted by prakharc

if y=0,z=0 x=50 and sum= 5000
x,y=0 z=25..sum=5000
x,z=0 y= 33..sum=5085...so contribution of y in sum is highest

So we get max sum if we maximize y......=> x=1,y=30,z=2..sum=4534

Though I thought of this solution when I saw these type of questions....don't have any formula or fixed method for such questions...so might be wrong

Answer is indeed 4534.

Is trail and error the only possible way for handling this question.

The answer given by TIME is as follows -

For maximum value of the expression is when y is maximised because the ratio of the squares of the coeffs of x,y,z in 2x+3y+4z=100 is 4:9:16 whereas the ratio of coeffs of x^2,y^2, z^2 in 2x^2 + 5y^2 + 8z^2 is 4:10:16. Therefore y^2 has relatively more weightage.

But i could nt understand this logic as well..
Can any one throw some light.

Thanks..

[ikruz]

Originally Posted by ahaprashant

Hello,
It's an AM>GM problem wherein you've to decide which possible combinations to take to form AM & GM and finally apply the inequality,
as in this case

(3x/2 + 3x/2 + 4y/3 +4y/3 +4y/3)/ 5 > (3x/2 x 3x/2 x 4y/3 x4y/3 x 4y/3) ^1/5 giving (3x + 4y) / 5 > (16 (x^2)(y^3)) ^1/5
OR, (3x + 4y) > 5 (16 (x^2)(y^3))^1/5 = 5(96)^1/5 = 10 (3)1/5
Hope that helps!

Pagal Gadha

Hi prasanth ..thanks..
but how did u arrive at 3x/2 and 4y/3 to start with?..Was it thru trial and error?.. or through intuition gained thru experience.?


In such questions, i am unable to get to this starting point of deciding these values...

[ahaprashant]

Originally Posted by ikruz

Hi prasanth ..thanks..
but how did u arrive at 3x/2 and 4y/3 to start with?..Was it thru trial and error?.. or through intuition gained thru experience.?


In such questions, i am unable to get to this starting point of deciding these values...

Not exactly trial n error,
Look at the terms in sum form: (3x + 4y)
Look at the terms in prod form: (x^2, y^3)
Looking at these, can we say we need to think of some X-term, two of which when multiplied offer an x^2 term with certain coefficient, and when summed up offer 3x as the sum, which suggests it shud be 3x/2. Similarly for Y case.

Hope that helps. (Btw its prashant )


Pagal Padha.