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From:

AMC 10 Q25 Solution

I get everything besides the last part. How in the world does he get:

$$3k + 2(867 - k) = 2013$$

I don't understand how he got this? What does this mean? Literally translated:

$$\text{3 times number of times we have 3 powers of 2 in between consecutive 5-powers}$$

$$\text{*Plus* 2 times the number of times we don't have 3 powers of 2 in between consecutive 5-powers. }$$

$$\text{Equals} \space 2013$$

How though?

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There are $2013$ powers of $2$ considered. In the $k$ highlighted rows, there are $3$ powers each and in the other, there are $2$, thus the list contains $2013 = 3k + 2(867-k)$ powers of two.

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  • $\begingroup$ Damn. That's all I got. $\endgroup$ – Amad27 Mar 4 '15 at 17:15
  • $\begingroup$ @Amad27 Huh? I don't understand your comment. $\endgroup$ – AlexR Mar 4 '15 at 17:16
  • $\begingroup$ I meant it's a great answer, simple, clear concise. This was in the interval, $(2^{2013}, 5^{867})$ right? What about in $(5^{867}, 2^{2014})$ $\endgroup$ – Amad27 Mar 4 '15 at 17:18
  • $\begingroup$ @Amad27 It was actually the interval $(1, 2^{2013})$. For the $2014$ case you need to check if $2^{2016} < 5^{868}$. If that's the case, you get one more triple for the extended interval and if not, the solution is the same as for $(1,2^{2013})$. $\endgroup$ – AlexR Mar 4 '15 at 17:21
  • $\begingroup$ Then what about $(2^{2013}, 5^{867})$? $\endgroup$ – Amad27 Mar 4 '15 at 17:25

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