:-)
Precisely, the very reason I'd put this question was b'coz there would be many who would try to make this complex when a little patience could have achieved quick results.

= 0
+ = (2+2^2)/3 - 1
+ = (2^3+2^4)/3 - 1

The trick is to spot the above pattern.

kool hai sire ..also tell me if my sequence one is rht or wrong and also how to calculate summation of the series when the diffrence of consecutive terms forms a g.p

:-)
Precisely, the very reason I'd put this question was b'coz there would be many who would try to make this complex when a little patience could have achieved quick results.

= 0
+ = (2+2^2)/3 - 1
+ = (2^3+2^4)/3 - 1

The trick is to spot the above pattern.

ohhh...i c.....that too kewl man .........

now there is nothing left in this problem .....thanks a lot man 4 putting such an eye opener question...

BTW: wot about my long approach ... is it correct ..or have i messed it up somewhere..???

0,0,1,2,5,10,21,42,85....

general term of the series ....... Tn + T(n-2) = 2^(n-3)
lets dividethe whole series based on positions in group of 2..
the sum of odd posiitons., So =(101st term -50) /3
the sum of even positions., Se =(100th term -9 /3
total sum = So+Se = (3*(101st term)-497) / 6 , considering 101st term = 2*(100th term) +1
Now general term of the series is.....Tn + T(n-2) = 2^(n-3)
so 101st term + 99th term = 2^98
nd100th term =4*(99th term )+1 -----------------------(2)
putting the value from two in total sum eqn..
Sum ={ (4^50 + 1)/10 } - 497/6

This is wot i am getting .................

:-)
Precisely, the very reason I'd put this question was b'coz there would be many who would try to make this complex when a little patience could have achieved quick results.

= 0
+ = (2+2^2)/3 - 1
+ = (2^3+2^4)/3 - 1

The trick is to spot the above pattern.

------------------------------------------------------------
Quant Question # 4
------------------------------------------------------------
Let denotes the greatest integer that is less than or equal to x, e.g. = 3, then + + + +...+ (101 terms) equals

(a) 1/3( 4^50 - 1) - 50 (b) 1/3( 4^51 - 1) - 51 (c) 2/3( 4^50 - 1) - 50
(d) none of the above

0,0,1,2,5,10,21,42,85....

general term of the series ....... Tn + T(n-2) = 2^(n-3)
lets dividethe whole series based on positions in group of 2..
the sum of odd posiitons., So =(101st term -50) /3
the sum of even positions., Se =(100th term -9 /3
total sum = So+Se = (3*(101st term)-497) / 6 , considering 101st term = 2*(100th term) +1
Now general term of the series is.....Tn + T(n-2) = 2^(n-3)
so 101st term + 99th term = 2^98
nd100th term =4*(99th term )+1 -----------------------(2)
putting the value from two in total sum eqn..
Sum ={ (4^50 + 1)/10 } - 497/6

This is wot i am getting .................

hi friends

iam having patni test on saturday.so please tell me solution for this puzzle.

A man leaves office daily at 7pm A driver with car comes from his home to pick him from office and bring back home

One day he gets free at 5:30 and instead of waiting for driver he starts walking towards home. In the way he meets the car and returns home on car He reaches home 20 minutes earlier than usual.

In how much time does the man reach home usually??

not meant to be posted here but man reaches home in 40 mins usually so if he gets free at 6 o clock then he reaches at 6:40

hi friends

iam having patni test on saturday.so please tell me solution for this puzzle.

A man leaves office daily at 7pm A driver with car comes from his home to pick him from office and bring back home

One day he gets free at 5:30 and instead of waiting for driver he starts walking towards home. In the way he meets the car and returns home on car He reaches home 20 minutes earlier than usual.

In how much time does the man reach home usually??

Ok...please take a pen and paper and solve the polynomials,
I've marked in bold, the end result of polynomials subtraction.

We have basically combined the terms from the result in such a manner to prove that
B^(x,y,z) - A(x,y,z) is always positive in both the scenarios when x>= 2z and when x<2z.

Let x > y > z > 0 and x >= 2z, then
B(x,y,z) = x+y-z,
if (B(x,y,z))^2 > A(x,y,z) => (B(x,y,z))^2 - A(x,y,z) > 0, solving we get
2y(3x-y) + 2x(3y-2z) which is > 0 as 3x-y > 0 and 3y-2z>0

Let x > y > z > 0 and x < 2z, then
B(x,y,z) = y+z+(2z-x) = 3z +y -x
if (B(x,y,z))^2 > A(x,y,z) => (B(x,y,z))^2 - A(x,y,z) > 0, solving we get
(3z +y -x)^2 - (x - 3y + 2z)(x-y) > 0 => 9z^2 - 2y^2 +8yz +2xy -8zx > 0
=> z(9z-8x+8y) + 2x(x-y) > 0, which we can see always is as
x>y, and 9z-8x+8y > 0(why?)

Got it sir.

9z-8x+8y>0. This is bcoz if we take boundary value for x i.e x=2z then expression becomes 8y-7z and y>z therefore, 9z-8x+8y>0. Anyways, x<2z hence value of the polynomial will always be positive.

Thanks a lot.

Right sir :-)

Solve and tell me.

1/24(1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 ) + 1/16(1/5 + 1/6)

bhai mere question ka bhi answer de do...woh diffrence is in gp ka summation wala

@aarav
bhai the nth term for the expression will be
tn = 1/(n-3)*(n-1)*(n+1)*(n+3)

apply partial fractions and seperate...uske baad sumation to infinity...rht or wrong ???

boss ths is like maggi hot n sweet,itz different,can't solve in tht way i fear,u will have 2 take (1,4),(2,3) combo 4 applyin partial fraction n separate,but thn..,:(
boss look wht aarav is tellin i.e +=2+2^2/3-1 identity

@aarav
bhai the nth term for the expression will be
tn = 1/(n-3)*(n-1)*(n+1)*(n+3)

apply partial fractions and seperate...uske baad sumation to infinity...rht or wrong ???

Your n starts from 6 then...be careful !

@aarav
bhai the nth term for the expression will be
tn = 1/(n-3)*(n-1)*(n+1)*(n+3)

apply partial fractions and seperate...uske baad sumation to infinity...rht or wrong ???

Right sir :-)

Solve and tell me.

crazyfootballer Says

I was not able to understand the solution. Got lost in between.

Ok...please take a pen and paper and solve the polynomials,
I've marked in bold, the end result of polynomials subtraction.

We have basically combined the terms from the result in such a manner to prove that
B^(x,y,z) - A(x,y,z) is always positive in both the scenarios when x>= 2z and when x<2z.

Let x > y > z > 0 and x >= 2z, then
B(x,y,z) = x+y-z,
if (B(x,y,z))^2 > A(x,y,z) => (B(x,y,z))^2 - A(x,y,z) > 0, solving we get
2y(3x-y) + 2x(3y-2z) which is > 0 as 3x-y > 0 and 3y-2z>0

Let x > y > z > 0 and x < 2z, then
B(x,y,z) = y+z+(2z-x) = 3z +y -x
if (B(x,y,z))^2 > A(x,y,z) => (B(x,y,z))^2 - A(x,y,z) > 0, solving we get
(3z +y -x)^2 - (x - 3y + 2z)(x-y) > 0 => 9z^2 - 2y^2 +8yz +2xy -8zx > 0
=> z(9z-8x+8y) + 2x(x-y) > 0, which we can see always is as
x>y, and 9z-8x+8y > 0(why?)

I'm surprised that no one has any doubts on last day's problem solution. I deliberately kept it non-verbose.
I hope the solutions are read by you all.
B^2(x,y,z) > A(x,y,z)?

I was not able to understand the solution. Got lost in between.

Yup option (c) is the answer.

Series goes like - 0+0+1+2+5+10+21+42+85+.....101 terms

I broke into 2 series - 1+5+21+.. 50 terms and 2+10+42+... 49 terms
Second series can be rewritten as 2(1+5+21+..) 49 terms

1+5+21.. can be solved by taking
a1=1
a2=4a1 + 1
a3=4a2 + 1 = 4(4a1 + 1) + 1
and so on

Solving for this series putting in the original series will give you option(c) as answer.

The method adopted by Apple and IdiotR is a better way i believe. In my method it can be little complicated.

Thank you for providing new and better ways of solving a problem. Purpose of this thread is being achieved (atleast for me).

Find the sum of following infinite series:
1/3.5.7.9 + 1/5.7.9.11 + 1/7.9.11.13 + ...

@aarav
bhai the nth term for the expression will be
tn = 1/(n-3)*(n-1)*(n+1)*(n+3)

apply partial fractions and seperate...uske baad sumation to infinity...rht or wrong ???

answer is (c).
Proof::
series is like
0 0 1 2 5 10 21 42 ......101 terms.
whenever we get odd number we multiply by 2 else (twice +1).
when we break it up ...
1 + 2 + 4 + 8 ........99 terms +
1 + 2 + 4 + 8 ........97 trerms +
..........................................
1+2+4 (3terms) +
1
using formula for gp for each series we get
2 + 2^3 + 2^5 +......+ 2^99 - 50 = option (c).
Cheers Apple
koi prob ho kisi ko to bolo bhailogo

yaar mein thoda detail mein solve kaarta hoon ok?
series
0 0 1 2 5 10 21 42 ......101 terms
sum=1+2+5+10+21+42+85+170+.....99 terms
=1+2+(4+1)+(8+2)+(16+5)+(32+10)+(64+21)+(128+42)+.....
=(1+2+4+8+16+32+64+128+......99 terms)+(1+2+5+10+21+42+85+170+....97
terms)
///Now break successive terms in similar way as in first step

=(1+2+4+8+16+32+64+128+......99 terms)+(1+2+4+8+16+32+64+128+......97
terms)+(1+2+4+8+16+32+64+128+......95 terms)+............+(1+2+5)
=(1+2+4+8+16+32+64+128+......99 terms)+(1+2+4+8+16+32+64+128+......97
terms)+(1+2+4+8+16+32+64+128+......95 terms)+............+(1+2+4)+1
= (2^99-1)+(2^97-1)+(2^95-1)+........+(2^3-1)+(2-1) (here used the (ar^n-1)/
(r-1) funda)
=2+2^3+2^5+2^7+..........+2^97+2^99-(1+1+1+...50 terms)
=something tht apple has calculatd-50
Answr option C

HAHA...was expecting it to be cracked before I could login for the day.
A 2 marker in CAT is expected to be solved in 150/180 seconds.

Those who still haven't found the logic to solve today's problem need not worry much.
Just look at this pattern
+ = (2+2^2)/3 - 1
and proceed.

In case you are free and need some glucose/horlicks for thought, try this:

Find the sum of following infinite series:
1/3.5.7.9 + 1/5.7.9.11 + 1/7.9.11.13 + ...

I'm surprised that no one has any doubts on last day's problem solution. I deliberately kept it non-verbose.
I hope the solutions are read by you all.
B^2(x,y,z) > A(x,y,z)?

Apple Says

uppar solve kara to samsite zara nazar uppar uthao n dekho :)

bhaisahab aapne detail mein nahi bataya kaise solve kiya woh sumation..mein calculation mein fasa hua huin..thoda time ho toh kripya detailed solution batayen..mujhe genral term aur summation nikalna sikhayaen...

@khana sir
jinke charanspash khud m.l khana karta hai kya woh khana sir mujhe apni problems jo unhone do din pehle post kari thi uska answer aur solution bateynge :D

regards
samsite

uppar solve kara to samsite zara nazar uppar uthao n dekho

hi bhailog..thodi help chahiye...

the series is like
0 + 0 + 1 + 2 + 5 + 10 + 21 + 42 + 85 + 170 + ......101 terms

this can be broken into

a) 1 + 5 + 21 + 85 + .......50 terms
and
b) 2 + 10 + 42 + 170 + .....49 temrs
the property of both the series is that the diffrence of the consequtive elements forms a g.p with a common multiplying factor of 4..
now b can be further reduced as 2*( series 1).
so what we get is
required sumation=3(1 + 5 + 21 + 85 + ....upto 50 terms) - last term of series 2.
i wanted to know how to write genral term for such expression and how to find the sum.

please tell me how to solve it as i have forgotten to solve for the sum of such types of series..
ciao
samsite

answer is (c).
Proof::
series is like
0 0 1 2 5 10 21 42 ......101 terms.
whenever we get odd number we multiply by 2 else (twice +1).
when we break it up ...
1 + 2 + 4 + 8 ........99 terms +
1 + 2 + 4 + 8 ........97 trerms +
..........................................
1+2+4 (3terms) +
1
using formula for gp for each series we get
2 + 2^3 + 2^5 +......+ 2^99 - 50 = option (c).
Cheers Apple
koi prob ho kisi ko to bolo bhailogo

The NL went out around 15 minutes ago. Lil teething issues .. surely to be sorted sooon!

------------------------------------------------------------
Quant Question # 4
------------------------------------------------------------
Let denotes the greatest integer that is less than or equal to x, e.g. = 3, then + + + +...+ (101 terms) equals

(a) 1/3( 4^50 - 1) - 50 (b) 1/3( 4^51 - 1) - 51 (c) 2/3( 4^50 - 1) - 50
(d) none of the above

Hmm just what i was looking for.....great work allwin!!!

answer is not (c) and a .
not c because
A(x,y,z) can be written as
(x-3y)^2 -6y^2 +2xy -2z(y-x)..
lets put x=0,y=1 and z=10 in this equation..we get -17 as answer..

@aarav
for the proof that A can take all the real values for the R set of x,y and z i put x=0 and y=1 and vary z from the limits -infinity to + infinity..one will get all the values.
regards
sameer

Good going guys, continue the spirit shown here so far :-)

The answer is choice (b).

Prove that A(x,y,z) can take any real value.
And give me a short and cleaner proof of B^(x,y,z) > A(x,y,z) for x>y>z>0.

now clearly this is always true, 4m > 4m-1

=> 4m - (4m-1) >0 --------(1)

but n although being lesser than m , can be a fraction such that ,

1>n>0 ...lets say n=1/2

in that
case 4mn=2m

then

4mn-(4m-1) = 2m -4m+1=1-2m <0 ...since m has to be greater than n that is greater than 1/2

hence ur 4mn-(4m-1)>=0 is violated here.....

so take the condition as m>n>=1 ..then it wud work fine :):)

Spot on dude,missd tht

meg_cat Says

w.r.t Quant Qtn 2.(x+y)(x-y) = 189....till here its fine but won't it be very difficult to find the factors ? (27*7)(1) = 63(3) = 27(7) = 21(9)any way out..?? help needed

Man ..... u cud have folowed the short cut to generate pythagorean triplets , but here 189 is not a perfect square ....

so we have to go like this

thats the way it is ....95*94 will give , y+x=95 nd x-y=94
63*3 will give , y+x=63 nd x-y=3
27(7) will give , y+x=27 nd x-y=7
21(9) will give , y+x=21 nd x-y=9

from each of the above eqns u can find the values of the pythagorean triplets .....givin 189 as the 3rd partner ....

then go ahead ..to find out the final soln ....

w.r.t Quant Qtn 2.(x+y)(x-y) = 189....till here its fine but won't it be very difficult to find the factors ? (27*7)(1) = 63(3) = 27(7) = 21(9)any way out..?? help needed

take x=mk,y=nk,z=k
m>n>o,k>0

4mn-(4m-1)>=0

now clearly this is always true, 4m > 4m-1

=> 4m - (4m-1) >0 --------(1)

but n although being lesser than m , can be a fraction such that ,

1>n>0 ...lets say n=1/2

in that
case 4mn=2m

then

4mn-(4m-1) = 2m -4m+1=1-2m <0 ...since m has to be greater than n that is greater than 1/2

hence ur 4mn-(4m-1)>=0 is violated here.....

so take the condition as m>n>=1 ..then it wud work fine :):)

from your mail

B(x,y,z) = x+y-z .......for x>2z
now B^2-A=-2y^2+z^2+6xy-4xz
take x=mk,y=nk,z=k
m>n>o,k>0

E=B^2-A =k^2=k^2
now frm m>n>o,k>0
2mn-2n^2>0 and 4mn-(4m-1)>=0
hence E>0
hence B^2>A
similarly for the other case i think we can prove the same
Also vineet no whr written in the prob tht x>y>z>0
it'z specific 2 option B
wat say u guyz?

ya rite..its specific to option b only...but it does nt matter any way..:D ..is nt it...:D

......nd ya ..similarly we cud prove the option c as u ve followed up on my prior explaination ....

for all values of x,y, z satisfying the condition given in problem
B(x,y,z) = x-x-y|| + |y-|y-z|| + |z-|z-x|| = y+z +|2z-x

B(x,y,z) = x+y-z .......for x>2z

B(x,y,z) = y-x+3z .......... for x<2z
...............


Also the expression x-y| + |y-z| + z-x = 2(x-z) =C for all values of x,y, z satisfying the condition given in problem

now 2x-2z >x+y-2z
C > B -z ..............for x>2z


so in that interval the solution (a) wont always hold good as z >= 1 , and for x<2z again it wont hold true ....

option c can be rulled out easily by puttin values ...

nd for all values of x,y, z satisfying the condition given in problem ,
the option b is holding ...

option (b) shud be correct

from your mail

B(x,y,z) = x+y-z .......for x>2z
now B^2-A=-2y^2+z^2+6xy-4xz
take x=mk,y=nk,z=k
m>n>o,k>0

E=B^2-A =k^2=k^2
now frm m>n>o,k>0
2mn-2n^2>0 and 4mn-(4m-1)>=0
hence E>0
hence B^2>A
similarly for the other case i think we can prove the same
Also vineet no whr written in the prob tht x>y>z>0
it'z specific 2 option B
wat say u guyz?

answer is not (c) and a .
not c because
A(x,y,z) can be written as
(x-3y)^2 -6y^2 +2xy -2z(y-x)..
lets put x=0,y=1 and z=10 in this equation..we get -17 as answer..

u cant take the values as x=0,y=1 and z=10


Its clearly stated in problem ....x > y > z > 0

samsite Says

sire if i prove the statement wrong for one triplet then the option will be wrong....so just did so for a speciic set of values of x,y and z.

boss dont take me otherwise,i meant tht whn u need 2 select 3 value combo 2 prove nethn wrng,itz very tough 2 come 2 a correct combo by trial n err,u had foung the right combo by selectin tht collectn,i wantd 2 know is tht by trial n err or nethn else
If tht hurt u ,i m xtremely sorry

Quant Question # 3
------------------------------------------------------------ Which of the following statements about the functions

A(x, y, z) = x + 3y - 4xy - 2yz + 2zx and B(x,y,z) = x-x-y|| + |y-|y-z|| + |z-|z-x| is necessarily true?

(a) B(x,y,z) <= x-y| + |y-z| + |z-x (b) (B(x,y,z))^2 > A(x,y,z) for x > y > z > 0

(c) A(x,y,z) > -6 (d) None of the above

for all values of x,y, z satisfying the condition given in problem
B(x,y,z) = x-|x-y|| + |y-|y-z|| + |z-|z-x|| = y+z +|2z-x

B(x,y,z) = x+y-z .......for x>2z

B(x,y,z) = y-x+3z .......... for x<2z
...............


Also the expression x-y| + |y-z| + z-x = 2(x-z) =C for all values of x,y, z satisfying the condition given in problem

now 2x-2z >x+y-2z
C > B -z ..............for x>2z


so in that interval the solution (a) wont always hold good as z >= 1 , and for x<2z again it wont hold true ....

option c can be rulled out easily by puttin values ...

nd for all values of x,y, z satisfying the condition given in problem ,
the option b is holding ...

option (b) shud be correct

sniff!!!sniffff!!!
i am getting this feeling:u hav editd the post
Also how u select ths 0,1,10 combo out of ths(-infi,+infi )range maannnnnnnnnnnnnnnnnnnn ???

Letz c wht guruji aarav decides

sire if i prove the statement wrong for one triplet then the option will be wrong....so just did so for a speciic set of values of x,y and z.

answer is not (c) and a .
not c because
A(x,y,z) can be written as
(x-3y)^2 -6y^2 +2xy -2z(y-x)..
lets put x=0,y=1 and z=10 in this equation..we get -17 as answer..

sniff!!!sniffff!!!
i am getting this feeling:u hav editd the post
Also how u select ths 0,1,10 combo out of ths(-infi,+infi )range maannnnnnnnnnnnnnnnnnnn ???

Letz c wht guruji aarav decides

Quant Question # 3
------------------------------------------------------------ Which of the following statements about the functions

A(x, y, z) = x + 3y - 4xy - 2yz + 2zx and B(x,y,z) = x-x-y|| + |y-|y-z|| + |z-|z-x| is necessarily true?

(a) B(x,y,z) <= x-y| + |y-z| + z-x (b) (B(x,y,z))^2 > A(x,y,z) for x > y > z > 0

(c) A(x,y,z) > -6 (d) None of the above

I am getting option (b) as answer. I did it by eliminating options (a) and (c). By putting some values in option (b) this is always coming true.

Not able to simplify the functions but will be testing my brain till the end of the day.

answer is not (c) and a .
not c because
A(x,y,z) can be written as
(x-3y)^2 -6y^2 +2xy -2z(y-x)..
lets put x=0,y=1 and z=10 in this equation..we get -17 as answer..

------------------------------------------------------------
Quant Question # 2
------------------------------------------------------------
The points P, Q, R lie on a line in that order with PQ=9, QR=21. Let O be a point not
on PR such that PO=RO and the distances PO and QO are integral.
Then sum of all possible perimeters of triangle PRO is

(a) 320 (b) 350 (c) 380 (d) 410

hey i saw this question today itself .....nd it seems easy to me ....


x any are the lengths of oq n op respectively,
we get (y-x)(y+x)=189=3*3*3*7 ,
since, we are asked for the integral values of y and x ,
we can have 4 poss values , of y and x respectively
33 , 30
15,6 ------ this does not satisfiy the property of triagle since here the sum of sides is equal to 3rd side.
17,10
95,94

so the sum of all the perimeters ....
33*2+30+17*2+30+95*2+30=66+34+190+90

=380 ....hence for integral values of x and y sum of perimeter s will be 380 :D

Ha d the condition of them being integral not given the answer wud have been infinite ...since there can be infinite triangles formed considerin the locus of the poin of the vertex to be the perpendicular bisector of pr.


expecting a good question today ...:D:D

Apple Says

(a) is definitely wrong put x=y=z (just an observation ...no funda) :)

yaaah surely

Quant Question # 3
------------------------------------------------------------ Which of the following statements about the functions

A(x, y, z) = x + 3y - 4xy - 2yz + 2zx and B(x,y,z) = x-x-y|| + |y-|y-z|| + |z-|z-x| is necessarily true?

(a) B(x,y,z) <= x-y| + |y-z| + z-x (b) (B(x,y,z))^2 > A(x,y,z) for x > y > z > 0

(c) A(x,y,z) > -6 (d) None of the above

is the ans option B??????

(a) is definitely wrong put x=y=z (just an observation ...no funda)

Quant Question # 3
------------------------------------------------------------ Which of the following statements about the functions

A(x, y, z) = x + 3y - 4xy - 2yz + 2zx and B(x,y,z) = x-x-y|| + |y-|y-z|| + |z-|z-x| is necessarily true?

(a) B(x,y,z) <= x-y| + |y-z| + z-x (b) (B(x,y,z))^2 > A(x,y,z) for x > y > z > 0

(c) A(x,y,z) > -6 (d) None of the above

Okay, this is probably the only solution I ever got on my own, so pls pls pls tell me this is correct.

Let (2^2)*(3^3)*(4^4)*(5^5)*(6^6)*(7^7) = K
And the other two sides be 'B' and hypotenuse 'H'

Then, by pythagoras,
K^2 = H^2 - B^2
= (H - B) (H + B) .... -(1)

Now, by triangle side sum property,
H + B > K and since H and B are integrals, and so is K,
H + B = K * i where i >=2 and is integer .... -(2)

From (1) and (2)
K ^2 = (H - B) (K * i)
K / i = (H - B) ..... -(3)

From (2) and (3)
H = (K*i^2 + K ) / 2*i and B = (K*i^2 - K )/ 2*i .... -(4)

For i = 2, 3,4,5,6,7,8,9 and 10 the sides are integral eg
for i = 11, H = 122K/11 and since K is not divisible by 11, H is not interal.

Hence there are 9 such triangles possible. Am I missing something or did I get something wrong again? ??:

PS: This is my first post here, so I don't know, but, am I suposed to wait some time for others to solve before I post?

U have not considered that i can be 2^2, 3^2,3^3, 2*3 etc. etc.

im late 2day........All questions solved.

Ok based on same funda

One of the smaller sides of a right angled triangle is
(2^2)*(3^3)*(4^4)*(5^5)*(6^6)*(7^7)
It is known that other two sides are integers.
How many triangles of this type are possible.

I believe i have put a real difficult problem..... Ok i make it easier change the length of the side to S = (3^3)*(5^5)*(7^7)(9^9)
Now it becomes much easier.

amar_kashyap Says

Okie, is it 31?...if yes i can post the solution..:)

no its not 31

im late 2day........All questions solved.

Ok based on same funda

One of the smaller sides of a right angled triangle is
(2^2)*(3^3)*(4^4)*(5^5)*(6^6)*(7^7)
It is known that other two sides are integers.
How many triangles of this type are possible.

Okay, this is probably the only solution I ever got on my own, so pls pls pls tell me this is correct.

Let (2^2)*(3^3)*(4^4)*(5^5)*(6^6)*(7^7) = K
And the other two sides be 'B' and hypotenuse 'H'

Then, by pythagoras,
K^2 = H^2 - B^2
= (H - B) (H + B) .... -(1)

Now, by triangle side sum property,
H + B > K and since H and B are integrals, and so is K,
H + B = K * i where i >=2 and is integer .... -(2)

From (1) and (2)
K ^2 = (H - B) (K * i)
K / i = (H - B) ..... -(3)

From (2) and (3)
H = (K*i^2 + K ) / 2*i and B = (K*i^2 - K )/ 2*i .... -(4)

For i = 2, 3,4,5,6,7,8,9 and 10 the sides are integral eg
for i = 11, H = 122K/11 and since K is not divisible by 11, H is not interal.

Hence there are 9 such triangles possible. Am I missing something or did I get something wrong again? ??:

PS: This is my first post here, so I don't know, but, am I suposed to wait some time for others to solve before I post?

im late 2day........All questions solved.

Ok based on same funda

One of the smaller sides of a right angled triangle is
(2^2)*(3^3)*(4^4)*(5^5)*(6^6)*(7^7)
It is known that other two sides are integers.
How many triangles of this type are possible.

Okie, is it 31?...if yes i can post the solution..:)

Finally able to crack wothout hit and trial.
Though of posting the solution But IdiotR has already given a detailed solution. So, i guess no need to post it again.

Guys see you tomorrow scratching your brain again.

Is there only ONE such trinagle..:)

As I'd said a lot many problems that appear in CAT can be traced from sources.
One such source is PGDCM paper of yester-years from IIM C.
The problem tomorrow is designed based on one question from PGDCM papers but with lot of modifications and innovations to build on more concepts from just 1 problem.

I will be off tomorrow b'coz of some personal work. You guys can carry on with the discussion.
A caveat: don't jump to an answer quickly

Please don't forget to see today's problem solution which talks about Stewart's theorem and its application.

im late 2day........All questions solved.

Ok based on same funda

One of the smaller sides of a right angled triangle is
(2^2)*(3^3)*(4^4)*(5^5)*(6^6)*(7^7)
It is known that other two sides are integers.
How many triangles of this type are possible.

In triangle OPR draw a perpendicular from 'O' to PR , OM.
Now we can write using pythogorous theorem,
OP^2-225=OQ^2-36
(OP+OQ)(OP-OQ)=189=3^3. 7...factorized into powers of prime factors. So there will be (3+1).(1+1)=8 factors in all and 8/2=4 ways in which the 189 can be expressed as a product of two factors
(Note: for any positive natural can be expressed as a product of powers of prime factors e.g, A=x^p.y^q... where x, y are prime nos. and the number of factors is =(p+1).(q+1)..and so on)
So, (OP+OQ)(OP-OQ)=189.1, 3.63, 9.21, 27.7, solving for different cases we will get tha values of (OP,OQ) as=(95,94),(33,30), (15,6), (17,10)...the case where OP=15 wont result into a trinagle becoz, in this case OP+OR=30=PR, which contradicts the llaw of triangle sum of two sides> any other side. Taking the other three cases we can find out all the possible triangles and their perimeters the sum of which will come out =380.

Is there any other way of doing this sum?..:)

to IdiotR,

Super solution man , I kept on trying with the sum of length of 2 sides gr8er than 3rd side funda and was getting nowhere.This was a trick i did not think of . Shall remember this solution , another suggestion , change ur username yaar:)

hi...need suggestions ..

--> Ravi, who lives in the countryside, caught a train for home earlier
than usual yesterday. His wife normally drives to the station to meet
him. But yesterday he set out on foot from the station to meet his
wife on the way. He reached 12 mins earlier than he would have done
had he waited at the station for his wife. The car travels at a
uniform speed, which is 5 times Ravi's speed on foot. Ravi reached
home exactly at 6 o' clock. At what time would he have reached home if his wife, forewarned of his plan, had met him at the station?

probable soln : D - distance from home to station.
d - distance between car and the station that day at the moment ravi arrives at the station.
x - speed of ravi on foot .
5x- speed of car .
t1- time taken for the car to reach station (pick him up) and return back to home.
= (d/5x + D/5x) (..in this case ravi waits for his wife at the station..)
t2-time taken for the car to meet ravi on the way (pick him up) and return back to home.
= (d/6x + (D-(d/6))/5x) (..in this case ravi has already started walking towards his home so he covers some distance before he meets his wife ..).

now t1 - t2 = 12/60 .
solving this we get d/x = 3 =>d=3x
if he had forewarned his wife about his plan , his wife had to either speed up or she had to leave for the station earlier to make sure that ravi does'nt ve to wait for her ..in order to do that she has to cover the distance 'd' ..i.e she has to save the time d/5x =3x/5x =36 mins.
Again had ravi waited for her on that day he would ve reached home by 6:12 .
thus now he reaches home at exactly 5:36.

................in this soln appropiate..if not soln. required...

Don't confuse it guys, I was mere answering a query.
The mirror image will still yield the same dimensions.
We are settled on this. Answer is (c) 380

If anyone of you has doubts on last problems solution we can take that up after some time.

@aarav
bhai time miley toh mere question ka answer de dena...
ciao
samsite

Apple Says

Yup Aarav no need to specify that we dont have to take mirror images ....we r working with magnitudes here. Wat say ??

Don't confuse it guys, I was mere answering a query.
The mirror image will still yield the same dimensions.
We are settled on this. Answer is (c) 380

If anyone of you has doubts on last problems solution we can take that up after some time.

Yup Aarav no need to specify that we dont have to take mirror images ....we r working with magnitudes here. Wat say ??

BINGO

PQ = 21, QR = 9, the dimensions of the triangle on the other side of the mirror would still be same. Try yourself and do let me know.

Hi aarav,i didn't understand this oneside of PR concept,we r workin with magnitude of sides,right?