official quant thread for cat08

[Svab]

Originally Posted by ahaprashant

Which means we're both close, post your approach, I guess probly I did some mistake somewhere. But I'm sure on 601 (Shouldn't be 7 for obvious reasons. Please share the approach.

Pagal Gadha

As u wrote that it will be 117^117 +118^118
we can also write it as 117^17 + 118^18
den first term will be :
117*117^16 or 117*689^8 0r 117*721^4 or 117*841^2 or
117*281= 877

second term : 924^9 or 924*776^4 or 924*176^2 or
924*976= 824

Sum of both terms = 877 + 824 = 701

[ahaprashant]

Originally Posted by vaibhav2cool

for the first equation
product of the roots=c

for the second equation product of the roots=c
so if one root is common then the other root also has to be common..

hence no value of b will give exactly one root common..
correct me if m wrong?

Hey I guess that's correct, but how does that justify the correctness of the answer viz. 4 (1 possible value of c)
Pagal Gadha

[mrholmes]

Originally Posted by masoom

see the motive is to use the inequality of AM and GM..

now.. since in product we have power of x as 5..

it means x is multiplying 5 times....

and power of y is 6...so y is multiplying 6 times...

so we have to break the part of x in 5 parts and y in 6 parts in sum ...

since we have to take limiting case.. all these terms would be equal..

hence 2x/5 and 3y/6..

thx masoom...

can we use same funda, AM > GM, for all such ques (whr sum is given & max product is asked OR product is given & min sum is asked)??


holmes

[ahaprashant]

Originally Posted by Svab

As u wrote that it will be 117^117 +118^118
we can also write it as 117^17 + 118^18
den first term will be :
117*117^16 or 117*689^8 0r 117*721^4 or 117*841^2 or
117*281= 877

second term : 924^9 or 924*776^4 or 924*176^2 or
924*976= 824

Sum of both terms = 877 + 824 = 701

Well, couldn't find anything wrong with your approach
(and hope your answer is correct)

Pagal Gadha.

[Vishal2704]

Originally Posted by vaibhav2cool

for the first equation
product of the roots=c

for the second equation product of the roots=c
so if one root is common then the other root also has to be common..

hence no value of b will give exactly one root common..
correct me if m wrong?

I guess no one able to approach the question correctly including me...

but somehow i've done it....correct me if i am wrong....

the only possible value that C can take is 0...for both the equations to have just one common root which is x=0 and for that b can take any number of values....apart from 14 in which case both the equations become same....

Waiting for ur replies puys...i guess i somehow crack the code

[ahaprashant]

Originally Posted by mrholmes

thx masoom...

can we use same funda, AM > GM, for all such ques (whr sum is given & max product is asked OR product is given & min sum is asked)??

holmes
holmes

Yes we can, but you need to be more specific as in which kind of problem are you referring to. But in general, wherever such sum-product problems appear, we can use this AM>GM>HM approach as long as one is able to realize / identify the possibility of the funda being used.

[Svab]

Originally Posted by ahaprashant

Well, couldn't find anything wrong with your approach
(and hope your answer is correct)

Pagal Gadha.

Its a very crude way to do but couldn find any other way to proceed..
Finding last t wo digits is much simpler.

If anyone finds a better way for last three digits,do let all puys know..

And how u got 601???
How u approached it...
Units and tens are ok but how hundred's?

[masoom]

Originally Posted by Vishal2704

I guess no one able to approach the question correctly including me...

but somehow i've done it....correct me if i am wrong....

the only possible value that C can take is 0...for both the equations to have just one common root which is x=0 and for that b can take any number of values....apart from 14 in which case both the equations become same....

Waiting for ur replies puys...i guess i somehow crack the code

this seems to be quite logically sound..

[ahaprashant]

Okay find out the last 2 digits of:
(i). 20674891 ^ 567
(ii). 61 ^ 8246

[ahaprashant]

Originally Posted by Svab

Its a very crude way to do but couldn find any other way to proceed..
Finding last t wo digits is much simpler.

If anyone finds a better way for last three digits,do let all puys know..

And how u got 601???
How u approached it...
Units and tens are ok but how hundred's?

Nope, there was a glitch in my approach for the 100th place digit.
701 is the answer for sure. Your approach tho' crude but I suppose that's the only way out