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I need help in solving this sequence:

24, 333333, ?, ?, 53, 44444444, 82, ?, ?, 222244444444, ?, ?, ?, 444433333322221

reposting this because I wanted to add more informations. Now by taking a look at the numbers that preceed 333333 and 44444444,we can see that 2+4=6 and after that we can see that 333333 is made of 6 numbers, the same goes for 5+3=8 we have 8 numbers 4 there.

Now take a look at 53, 44444444, 82 you can see that the number that preeceds 44444444 is 82-30+1=53 so i just assumed that for the next number we should use 82+30-1=111 11+1=12 222244444444 has 12 numbers and the criteria starts working now we do the same 111+30-1=140 and we put it right after 222244444444 ,then we do again 140+30-1 we firstly do 140+30=170 now we subtract -1 from 0 and for some reasons it becomes 16-1 now 16-1=15 and 444433333322221 has 15 numbers so by substituting these values from the original sequence we get this one:

24, 333333, K, ?, 53, 44444444, 82, ?, 111, 222244444444, 140, ?, 16-1, 444433333322221

I'm unsure about k,because by using my criteria that should have been 24+30-1=53 but we find that later before 44444444. The only thing i know is that after 82 we need a 10 digit number,because of 8+2=10, after 140 we need a 14 digits number please help me

source:http://zagaza.ru/za548.htm author has autorized repost of this

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  • $\begingroup$ you could see 3333 ,44444 etc as a sum,Multiplication,division,subraction of those values etc $\endgroup$ – alnesi Jan 13 at 14:28
  • $\begingroup$ @Alconja The link must have been removed. $\endgroup$ – rhsquared Jan 14 at 7:51

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