
Brain Teaser #2
5.101001000100001....
he decimal number aboce consists of only one and 0 to the right of the decimal point. The first is followed by one 0, the second 1 is followed by two 0, the third 1 is followed by three 0 and so on. What is the total number of 0s between the 98th and the 101st 1 in this decimal number?
what i supposed was that 98th 1 has 98 0, and so on, thus i added up 98+99+100=297.... but for some reason the asnwer is 291. I need a good explanation, and some tips to solve number pattern problems and sequence problems. Thank you very much for your help!

Hello,
The nth 1 is preceded by a number of 0 equal to $\displaystyle S_n=1+2+\dots+(n1)=\sum_{k=1}^{n1} k=\frac{n(n1)}{2}$
The pth 1 is preceded by a number of 0 equal to $\displaystyle S_p=1+2+\dots+(p1)=\sum_{k=1}^{p1} k=\frac{p(p1)}{2}$
Hence, assuming that p>n, there is a number of 0 between the nth and the pth equal to :
$\displaystyle S_pS_n=\frac 12 (p(p1)n(n1))$
This is the theorical part..
Here, p=101 and n=98... And I find the same result :)