IIT JEE , AIEEE Forum - Mathematics , Algebra - Try this

sandeepramesh

X plays the following game.
He generates a sequence where he initially puts down (deliberately) the values 1 and 4 as the 1st 2 terms of the sequence.
The rules of the game are :
1. Whenever you have a perfect square which is deliberately put down by you as the last term of a sequence, then the next term of the sequence is got by adding it and the previous term of the sequence.
2. If the last term of the sequence is a non perfect square
Case 1 : not divisible by 3
Put deliberately the next perfect square of the sequence (next to the last perfect square in the sequence) and follow step1.
Case 2 : divisible by 3
Subtract 5 to get the next term of the sequence
3. If the last term of the sequence is a perfect square got by 2 (i.e. not deliberately added) then add to the sequence the next perfect square and follow step 1.
For Example the sequence proceeds as :
1,4
1,4,5
1,4,5,9
1,4,5,9,14
1,4,5,9,14,16
1,4,5,9,14,16,30
1,4,5,9,14,16,30,25
1,4,5,9,14,16,30,25,36
1,4,5,9,14,16,30,25,36,61
1,4,5,9,14,16,30,25,36,61,49
1,4,5,9,14,16,30,25,36,61,49,110
1,4,5,9,14,16,30,25,36,61,49,110,64
1,4,5,9,14,16,30,25,36,61,49,110,64,174
1,4,5,9,14,16,30,25,36,61,49,110,64,174,169
...............................................................

The question is : While playing such a game, how many perfect squares does X miss till 1,00,00,000. Razz

sandeepramesh

Ummm if this would inspire many guys n gals here

Rates assured! Razz

sboosy

some of the first few terms of this sequence are
1,4,5,9,14,16,30,25,36,61,49,110,64,174,169,196,365,225,590,256,846,841,900,1741,961,2702,1024,3726,3721,3844,7565,3969,11534,4096,15630,15625, .....

now notice that every 7th term is forced to be reduced by 5 ...
which tells us there is some pattern .....

now further checking brings us this beautiful result
leave the first 8 squares ....
the squares of 9 ,10 , 11 , 12 are omitted ...and then the next four get included ....
then again the squares of (17,18,......28) get omitted ....and the next four are included
then again the squares of (33,34,.....60) get omitted ..and the next four are included .
then again the squares of (65,64,65,....124) get omitted and the next 4 included ...
the pattern goes like this
first 4 squares get omitted(9 ,10 , 11 ,12) (this ends with 12)
next after leaving 4 squares untouched
12 squares get omitted (17 to 28) (this ends with 28)
now after leaving another 4 untouched
28 squares get omitted (33 to 60) (this ends with 60)
now after this leaving another 4 untouched
60 squares get omitted(65 to 124) ...
thus
the missed squares follow sequence ...
we get
4 , 12 , 28 , 60 , ......
and the square just below 10000000 is 3162 square ...
so we shud see which number close to that falls in this sequence ..
and sum it upto that
it is easy to sum it up as we see ..that their common differences are in GP ...but slightly long ..
i ll try to post the exact answer in sometime

elastiboysai

4,12,28,60,124...sumthin like dat??
till u reach the no closest to 3162
now sum dat up Razz

sandeepramesh

what's the genl term of the no. of missing nos sequence?

i.e. 4,12,28, etc etc?

That was my method exactly but go on n post the answer like how to go abt calculating?

akhil_o

is it
4, 4(1+2), 4(1+2+4), 4(1+2+4+8)...and so on?
like 4(1+2+4+...2^n-1)?

sandeepramesh

why so long?

its 4(2^n-1) Razz

sboosy

the answe r is
4[ (2-1)+(4-1)+(8-1) ......(256-1)]
4[510 - 8]
4*(502) = 2008 ..
pls correct if wrong

sandeepramesh

i have to find it myself Razz

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