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Let $\displaystyle N = \sum^{2009}_{r=1}\bigg \lfloor \frac{2^r}{5}\bigg \rfloor$, where $\lfloor x\rfloor $ is a floor function of $x$. I want to find the remainder when $2^{2010}$ is divided by $N$.

I have used $\lfloor x \rfloor = x-\{x\}$, where $\{x\}$ represents the fractional part of $x$.

So $$\displaystyle N = \sum^{2009}_{r=1}\bigg \lfloor \frac{2^r}{5}\bigg \rfloor = \sum^{2009}_{r=1}\frac{2^r}{5}-\sum^{2009}_{r=1}\bigg\{\frac{2^r}{5}\bigg\} = \frac{1}{5}\bigg(\frac{2^{2010}-2}{1}\bigg)-\sum^{2009}_{r=1}\bigg\{\frac{2^r}{5}\bigg\}$$

I want be able to go further, could someone help me with this? Thanks.

1 Answers 1

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What you need is the remainder of $2^r$ modulo $5$. Since $2^4 = 16 = 3\cdot 5 + 1$, that remainder is periodic with period $4$, and each possible nonzero remainder modulo $5$ occurs, $2^0 \equiv 1\pmod{5}$, $2^1\equiv 2 \pmod{5}$, $2^2 \equiv 4 \pmod{5}$, $2^3 \equiv 3 \pmod{5}$. So the sum of any four consecutive remainders of $2^r$ modulo $5$ is $1+2+3+4 = 10$. In

$$\sum_{r = 1}^{2009} \biggl\{\frac{2^r}{5}\biggr\},$$

we have one instance of $\frac{2}{5}$ followed by $\frac{2009 - 1}{4} = 502$ groups of four consecutive remainders, so

$$\sum_{r = 1}^{2009} \biggl\{\frac{2^r}{5}\biggr\} = \frac{2 + 502\cdot 10}{5} = \frac{5022}{5}.$$

Then we have

$$N = \sum_{r = 1}^{2009} \biggl\lfloor \frac{2^r}{5}\biggr\rfloor = \frac{\bigl(2^{2010} - 2\bigr) - 5022}{5} = \frac{2^{2010} - 5024}{5},$$

and

$$2^{2010} - 5N = 5024.$$