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Originally Posted by grimreaper DUde.... sum of "squares " of 3 consecutive odd numbers
hehehe...try again oops sorry,
Op3)5555
the 3 nos are 41,43,45
CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

12n(n+1)+11 = Sum
Only 555511=5544 is div by 12
Also,then n=21
Hence the nos are 41,43,45
CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Originally Posted by shaunak_87 oops sorry,
Op3)5555
the 3 nos are 41,43,45 Let the numbers be n4 , n2 ,n
so i get,
(n4)^2+(n2)^2+n^2 = 5555
so
Solving this i found n value to be 45.
Some1 can post efficient method than this if exists 

Hedonist
Originally Posted by madhu89 Let the numbers be n4 , n2 ,n
so i get,
(n4)^2+(n2)^2+n^2 = 5555
so
Solving this i found n value to be 45.
Some1 can post efficient method than this if exists well how did u take it to be 5555 in the first place... ?
n is least of my concern...... post the method as to how you got 5555 as the answer .... cause if you get 5555 u get the numbers too... 

let the no be n2 , n ad n+2
now (n2)^2 + (n)^2 + (n + 2)^2 = 3n^2 + 8
Now only one options satisfies the condition. Hence 5555

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Hedonist
Originally Posted by aloknitw let the no be n2 , n ad n+2
now (n2)^2 + (n)^2 + (n + 2)^2 = 3n^2 + 8
Now only one options satisfies the condition. Hence 5555 correct
i hope u applied logic to select 5555 and not checking each number one by one......

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@GrimSaar
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CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Hedonist
Originally Posted by shaunak_87 @GrimSaar
Some tips to improve Speed lol naam ki rape kardi be :P :P....
where are the mods by the way?? i mean... long time..havent seen them online at all......
and well hope more ppl start visiting the forum... its really suna suna for now 

Hedonist
official solution for Q3 (my method)
Originally Posted by grimreaper Lets number the Question ok.... so that way its easier to keep track..... Q3) N is the sum of squares of 3 consecutive odd numbers such that all digits of N are same. If N is a four digit no. then N is 
a)9999
b)7777
c)2222
d)5555
also what are the three consecutive numbers ?
p.s when someone posts a good solution feel free to show yo appreciation by clicking on the thanks button......
well, first of all let the numbers be n2,n and n+2
so sum of squares would be (n2)^2 + n^2 + (n+2)^2 = 3n^2+8
now see the options...
only two numbers that is 2222 and 5555 when 8 is sub are divisible by 3
that is 2214 and 5547,
when divided by 3 we get... 738 and 1849
clearly we knw that a square can never end in 8 hence 1849 is the answer...
which is nothing but 43^2
hence your answer... 41,43,45 
The Following 7 Users Say Thank You to grimreaper For This Useful Post:
jammy99 (01Nov11), pratty89 (23Sep11), rahulsah1987 (17Jul11), rgrohitgrover (01Aug11), shaunak_87 (29Mar11), shonk (31Mar11), sreekanthg (29Oct11)

Originally Posted by grimreaper Lets number the Question ok.... so that way its easier to keep track..... Q3) N is the sum of squares of 3 consecutive odd numbers such that all digits of N are same. If N is a four digit no. then N is 
a)9999
b)7777
c)2222
d)5555
also what are the three consecutive numbers ?
p.s when someone posts a good solution feel free to show yo appreciation by clicking on the thanks button...... Let the three consecutive odd numbers be X2, X and X+2.
Now given that N= (X2)^2 + X^2+(X+2)^2
N= X^2 + 44X +X^2 + X^2 + 4 + 4X
=> N= 3X^2 + 8
If we put N=1111 then
N=3X^2+8
=> 3X^2+8 =1111
=> 3X^2=11118
X^2= (n8)/ 3 (1)
The values of N which are not divisible by 3 will be eliminated since we want the values of X as positive odd integers.
We will substitute values of N one by one now and see which one satisfes the equation (1)
Options (a),(b) don't satisfy it.
If N= 2222 then
X^2 =(22228)/3
= > 2214/3
X^2 = 738
But since we know that a perfect square can't end in 8
so N can't be 2222
if N=5555 then
X^2=(55558)/3
= > X^2 = 1849
so X= 43
Now X2 = 432= 41
and X+2 = 43+2 = 45
thus the three consecutive odd integers are 41, 43 and 45
so N= (41)^2 + (43)^2 + (45)^2
= > 1681 +1849 + 2025
hence N = 5555

The Following 3 Users Say Thank You to anjalig For This Useful Post:
aman99 (10Feb13), exams.2011 (27Jul11), grimreaper (29Mar11)

Originally Posted by grimreaper
well, first of all let the numbers be n2,n and n+2
so sum of squares would be (n2)^2 + n^2 + (n+2)^2 = 3n^2+8
now see the options...
only two numbers that is 2222 and 5555 when 8 is sub are divisible by 3
that is 2214 and 5547,
when divided by 3 we get... 738 and 1849
clearly we knw that a square can never end in 8 hence 1849 is the answer...
which is nothing but 43^2
hence your answer... 41,43,45
hi
only two numbers that is 2222 and 5555 when 8 is sub are divisible by 3
what do you mean by " when 8 is sub "
please explain


Hedonist


Originally Posted by grimreaper oh... when 8 is subtracted.... sorry about the typo oh thanks !
btw what about my solution
is there any other approach to solve this ques.


Hedonist

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Hedonist
Q 4
Q4) 3*2^p + 5^q + 7^r + 11^s =2008 what is the value of p+q+r+s, where all are greater than 0
a)9
b)10
c)24
d)15
e)unique solution does not exist 

Originally Posted by grimreaper Q4) 3*2^p + 5^q + 7^r + 11^s =2008 what is the value of p+q+r+s, where all are greater than 0
a)9
b)10
c)24
d)15
e)unique solution does not exist op e)unique solution does not exist
p q r s
0 4 2 3
9 0 3 2
Last edited by shaunak_87; 29Mar11 at 2:37 PM.
CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Hedonist
Originally Posted by shaunak_87 op e)unique solution does not exist
p q r s
0 4 2 3
8 0 3 2 the second set.... 8 0 3 2 where does that satisfy the condition ???
how did you get the second set O.O
am i missing something ??? O.o


Originally Posted by grimreaper the second set.... 8 0 3 2 where does that satisfy the condition ???
how did you get the second set O.O
am i missing something ??? O.o edited the post
9 0 3 2
Made a mistake had calculated 512 = 2^8 CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Hedonist
Originally Posted by shaunak_87 edited the post
9 0 3 2
Made a mistake had calculated 512 = 2^8 then also its coming upto only 2001
i mean how can there be another pair ?? i mean the very basis of the solution is that only one pair can exist.... think think u making a silly mistake 

Originally Posted by grimreaper then also its coming upto only 2001
i mean how can there be another pair ?? i mean the very basis of the solution is that only one pair can exist.... think think u making a silly mistake True
Just 1 possible, peeking a boo in office and trying to increase speed...and accuracy goes for a toss...
That's y, how to be accurate and increase speed?? CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 
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