Solve $\lfloor{x}\rfloor+\lfloor2x\rfloor+\lfloor4x\rfloor+\lfloor16x\rfloor+\lfloor32x\rfloor=12345$

Question

Solve for $x$
$$\lfloor{x}\rfloor+\lfloor2x\rfloor+\lfloor4x\rfloor+\lfloor16x\rfloor+\lfloor32x\rfloor=12345$$

I tried to put $x$=$I$+$f$ where $I$ is integer part and $f$ is fractional part but that didn't work.

Answer 1

As $y-1 < \lfloor y \rfloor \leq y$ you get

$$x+2x+4x+16x+32x -5< \lfloor{x}\rfloor+\lfloor2x\rfloor+\lfloor4x\rfloor+\lfloor16x\rfloor+\lfloor32x\rfloor \\ \leq x+2x+4x+16x +32 x$$

Thus

$$55x-5 < 12345 \leq 55x$$

or

$$11x-1 < 2469 \leq 11x$$

This implies

$$ 224.\bar{45} \leq x \leq 224.\bar{54}$$

Now all you need is to see the intervals for $2x, 4x, 16x, 32x$ and see how many choices you have for each.

Answer 2

Let $$f(x)=\lfloor x\rfloor+\lfloor 2x\rfloor+\lfloor 4x\rfloor+\lfloor 16x\rfloor+\lfloor 32x\rfloor\;.$$

By actual calculation $f(224)=55\cdot224=12320$ and $f(224.5)=12347$. Suppose that $224<x<224.5$, say $x=224.5-\epsilon$. Then

$$\begin{align*} f(x)&=224+\lfloor 449-2\epsilon\rfloor+\lfloor 898-4\epsilon\rfloor+\lfloor 3592-16\epsilon\rfloor+\lfloor 7184-32\epsilon\rfloor\\ &=(224+449+898+3592+7184)+\lfloor-2\epsilon\rfloor+\lfloor-4\epsilon\rfloor+\lfloor-16\epsilon\rfloor+\lfloor-32\epsilon\rfloor\\ &=12347+\lfloor-2\epsilon\rfloor+\lfloor-4\epsilon\rfloor+\lfloor-16\epsilon\rfloor+\lfloor-32\epsilon\rfloor\\ &\le 12343\;, \end{align*}$$

since $\lfloor -k\epsilon\rfloor\le-1$ for $k>0$. Thus, the equation $f(x)=12345$ has no solution.

Answer 3

No non-integer rational or real numbers are needed to solve this.

Write $i=\lfloor x\rfloor$ the integral part, and $f=x-i$, the fractional part of $x$. Since $\lfloor nx\rfloor=ni+\lfloor nf\rfloor$, the contibutions of $i$ and $f$ to the left hand side can be separated, and are $(1+2+4+16+32)n=55n$ and $\lfloor 2f\rfloor+\lfloor 4f\rfloor+\lfloor 16f\rfloor+\lfloor 32f\rfloor$ respectively (one has $\lfloor 1f\rfloor=0$). Since the latter is always less than$~54$ (in fact it is at most$~50$), it is clear that $i=12345\div55=224$ is a necessary condition for a solution (where integer division is written as $a\div b=\lfloor a/b\rfloor$), and what remains is $\lfloor 2f\rfloor+\lfloor 4f\rfloor+\lfloor 16f\rfloor+\lfloor 32f\rfloor=12345\bmod 55 = 25$.

Now putting $m=\lfloor 32f\rfloor$, this can be written as $$ m\div16+m\div8+m\div2+m=25.\tag1 $$ Writing the integer $m$ in base$~2$, five bits suffice since $m<32$. Each bit position has a separate contribution to (1) since integer division by a power of$~2$ is a right-shift on the bits, dropping the least significant one. The contributions of the bits, when equal to$~1$, are $1+2+8+16=27$ for the bit in position$~4$ (the one representing $2^4=16$), then $27\div2=13$ for the bit in position$~3$, and respectively $13\div2=6$, $6\div2=3$ and $3\div2=1$ for the remaining positions $2,1,0$. Now clearly $27$ is too much to attain $25$, but the sum $13+6+3+1=23$ of the remaining possible contributions is too little. So $(1)$ has no (integer) solutions for $m$, and there are no solutions to this problem.

One can easily deduce from this a general method to solve this kind of problems using only integer arithmetic, provided the integer factors$~n$ in the terms $\lfloor nx\rfloor$ each divide their successor. If they are all powers of a fixed base$~b$ as is the case here, one expresses $m$ in base$~b$ and everything works as before. For the general case one needs a mixed-radix representation of$~m$, chosen such that each digit position contributes to a sequence of final terms in the equation corresponding to$~(1)$. For instance to solve $$ m\div105+m\div15+m\div5+m=444,\tag2 $$ one chooses mixed radix with place values $1,5,15,105$, where $m=105d+15c+5b+a$ with $a<5$, $b<3$, $c<7$ contributes $134d+19c+6d+a$. Those coefficients were computed as from right to left as $$\begin{align} 1/1&=1,\\5/1+5/1=1+5&=6=3\times1+1,\\15/15+15/5+15/1=1+3+15&=19=3\times6+1\\ 105/105+105/15+105/5+105/1=1+7+21+105&=134=7\times19+1.\end{align} $$ A greedy algorithm solves the the knapsack problem with $d=3,c=2,b=0,a=4$ so $(2)$ has the unique solution $m=349$.