Is there a good way to solve this problem?

Find the range of $k\in \Bbb R$ so that there are 16 integer points ${(a,b)}$ that exist on or within the region enclosed by (including boundaries) $y=2^{x-4}$ and $y=\log_2 x + k$.

This was a question given to high school students. I am asking because I couldn't find a nice way to solve it except maybe some dirty labor (even with some dirty labor, I couldn't find the answer yet). But I see this type of questions over and over again. So I was wondering if the guy who made this question (probably a math teacher in some school) had a nice solution.


closed as off-topic by user296602, Shailesh, Leucippus, user186473, user91500 Aug 25 '16 at 3:52

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  • $\begingroup$ Are we finding $p$ and $q$ or the number of integer points? $\endgroup$ – Simply Beautiful Art Aug 24 '16 at 22:31
  • $\begingroup$ @SimpleArt The range of $k$. I revised to clarify. $\endgroup$ – Kay K. Aug 24 '16 at 22:38
  • $\begingroup$ with some dirty labor I found $k\in[2,5-\log_2 6)$ $\endgroup$ – Doug M Aug 24 '16 at 23:19
  • $\begingroup$ @DougM Thanks. Could you share how you solved it? Dirty labor may be the only way. $\endgroup$ – Kay K. Aug 24 '16 at 23:26

Let's call $k$ admissible if the corresponding curves enclosed exactly 16 lattice points. It is clear that the number of enclosed lattice points is a monotonically increasing function of $k$. Let $f(x) = 2^{x-4}$ and $g(x) = \log_2{x}+k$. Suppose first that $k$ is an integer. For a given integer $a$, let $h(a)$ be the number of enclosed lattice points with $x$-coordinate $a$. If $k$ is positive, we note that \begin{align} f(1) &= \frac{1}{8}, g(1) = k\implies h(1) = k \\ f(2) &= \frac{1}{4}, g(2) = k+1 \implies h(2) = k+1 \\ f(3) &= \frac{1}{2}, g(3) = k+\log_2{3} \implies h(3) = k+1 \\ f(4) &= 1, g(4) = k+2 \implies h(4) = k+2 \\ f(5) &= 2, g(5) = k+\log_2{5} \implies h(5) = k+1 \\ f(6) &= 4, g(6) = k+\log_2{6} \implies h(6) = \max(k-1,0) \\ f(7) &= 8, g(7) = k+\log_2{7} \implies h(7) = \max(k-5,0). \end{align} Note that for $a\ge 7$, we have $f(a+1)-f(a)>1$ and $g(a+1)-g(a)<1$, so $f$ grows much faster than $g$ for $a\ge 7$, and so $h$ would decrease if it weren't already zero.

In particular, for $k=2$ we have $h(a)>0$ for $1\le a\le 6$, and the number of enclosed lattice points is $$ 2+ (2+1) + (2+1) + (2+2) + (2+1) + (2-1) = 16. $$ So $k=2$ is an admissible value. Furthermore, for $k=2$ we have lattice points lying on the boundary of the upper curve $g$ (e.g. $(1,2)$, $(2,3)$, $(4,4)$). Thus, if $k$ is lowered to $2-\epsilon$ for any $\epsilon>0$, then these points will no longer be enclosed, and so the lowered value of $k$ would not be admissible. Thus, $k=2$ is the minimum of the admissible set.

Now, we ask how much we can increase $k$ to $2+\epsilon$ so that we still enclose 16 lattice points. Since for $k=2$ we don't have boundary points on the upper curve for $a=3$, $5$, and $6$, we can translate the curve up until it hits a lattice point for one of the preceding values of $a$. The maximum $\epsilon$ we can raise $k$ is thus the smallest of the vertical distances between $g(a)$ for $a=3,5,6$ and the next lattice points. The distance between $g(a)$ and the next highest lattice point for $a=3,5,6$ are $$ 2-\log_2{3}, 3-\log_2{5}, 3-\log_2{6} $$ respectively. These distances are equal to $\log_2{4/3}$, $\log_2{8/5}$, and $\log_2{8/6} = \log_2{4/3}$, respectively. The smallest of those quantities is $\log_2{4/3}$. Thus, $k=2+\epsilon$ for any $0<\epsilon<\log_2{4/3}$ is admissible, while $k = 2+\log_2{4/3}$ is not admissible.

Thus, the set of admissible $k$ is $\boxed{[2,2+\log_2{4/3})}$.


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