How many solutions?

• If a solution exists , x must be positive:

If x < 0, then:

$\left\lfloor ax \right\rfloor < 0 \hspace{5 mm} \forall a \in \mathbb{N}$

so : $\left\lfloor x \right\rfloor + \left\lfloor 2x \right\rfloor + \left\lfloor 4x \right\rfloor + \left\lfloor 4x \right\rfloor + \left\lfloor 8x \right\rfloor + \left\lfloor 16x \right\rfloor + \left\lfloor 32x \right\rfloor < 0 \neq 12345$

Therefore, for a solution to exist, x must be positive .

• Assume that a solution exists:

If a solution exists, it will be in the form:

$x = n + p \hspace{5 mm} : n \in \mathbb{N}, \hspace{1 mm} p \in \mathbb{R}, \hspace{1 mm} 0 \le p < 1$

such that:

$\left\lfloor x \right\rfloor = n\hspace{2 mm},$

$\left\lfloor 2x \right\rfloor = \left\lfloor 2(n + p) \right\rfloor = \left\lfloor 2n + 2p \right\rfloor \in \left\{ 2n, 2n + 1 \right\} \hspace{2 mm} (\text{because } 0 \le 2p < 2) \hspace{2 mm},$

$\left\lfloor 4x\right\rfloor = \left\lfloor 4n + 4p \right\rfloor \in \left\{ 4n, 4n + 1, 4n + 2, 4n + 3\right\} \hspace{2 mm} (\text{because } 0 \le 4p < 4) \hspace{2 mm},$

$\left\lfloor 8x \right\rfloor \in \left\{ 8n + i : i \in \mathbb{N}, i \le 7\right\}\hspace{2 mm},$

$\left\lfloor 16x \right\rfloor \in \left\{ 16n + i : i \in \mathbb{N}, i \le 15\right\}\hspace{2 mm},$

$\left\lfloor 32x \right\rfloor \in \left\{ 32n + i : i \in \mathbb{N}, i \le 31\right\}$

And this gives us that: $\left\lfloor x \right\rfloor + \left\lfloor 2x \right\rfloor + \left\lfloor 4x \right\rfloor + \left\lfloor 4x \right\rfloor + \left\lfloor 8x \right\rfloor + \left\lfloor 16x \right\rfloor + \left\lfloor 32x \right\rfloor = 63n + m \hspace{5 mm} : m \in \mathbb{N}$

The 63n comes from adding (n + 2n + 4n + 8n + 16n + 32n) that results from the floor operation. m is the integer that results from adding up all the integer terms that come out of the floor operations (like 3 in 4n + 3 if $\left\lfloor 4x \right \rfloor$ happens to equal 4n + 3)

A lower bound on m is found when no integer terms come out of the floor operations. In this case, m = 0

An upper bound on m is found when every floor operation results with the maximum integer term. In this case, m = 0 + 1 + 2 + 3 + 7 + 15 + 31 = 59

So now we know that : $0 \le m \le 59$

Now, the equation at hand becomes: $63n + m = 12345$

$\therefore 63n = 12345 - m$

$\therefore n = \frac{12345 - m}{63}$

and becuase n must be an integer, and 12345 is not divisible by 63. Then m must be the reminder of 12345/63 .

$\therefore m = 12345 \text{ mod } 63 = 60$

$\therefore$ for a solution to exist, m must be equal to 60.

but we have : $0 \le m \le 59$

$\therefore$ m cannot equal 60.

$\therefore$ There is no solution.

$\therefore$ The sum of all possible solutions is 0

answer to the problem is given in the "NOTE" itself

Umm..don't you think that the 'Note' in the question kind of gives away the answer. I simply read the note & typed 0 in the answer box.

$\lfloor{x}\rfloor$ + $\lfloor{2x}\rfloor$ +...+ $\lfloor{32x}\rfloor$ =12345

CASE 1:- Let $x\in\Z$

Then, $63x=12345$

Obviously, no integer $x$ satisfies the given equation.

CASE 2:- Let $x\notin\Z$

Then, rewriting $\lfloor{x}\rfloor$ as $x-\{x\},$

$63x-63\{x\}=12345$

We get $63\lfloor{x}\rfloor=12345$

As $\lfloor{x}\rfloor$ must be an integer, no value satisfies the given equation.

Hence number of values of $x$ satisfying the equation come out to be $\boxed{0}$

If $x$ is an integer, we know the left-hand side will equal to 12348 when $x=196$ , so we put down the value of $x$ a little bit such that we can just minus out $3$ , but since we drop down the value of $x$ and the six floor will all minus at least $1$ , the smaller value it can get is $12342$ , it is not possible to get $12345$ , just answer $\boxed0$ then you're done!

I need you guys to tell me if what I did was right: first, for all integers $n$ between $0$ and $5$ it is true that $\lfloor 2^n x \rfloor \leq 2^n x < \lfloor 2^n x \rfloor + 1$ . Then, if we sum

$\sum_{n=0}^5 \lfloor 2^n x \rfloor = 12345 \leq \sum_{n=0}^5 2^n x = 63x < \sum_{n=0}^5 (\lfloor 2^n x \rfloor + 1) = 12360,$

then $195.952381 \leq x < 196.1904762$ . But then again, as $\lfloor 2^n x \rfloor \leq 2^n x,$ there must be some $x$ and $n$ such that

$\frac{ \lfloor 2^n x \rfloor }{2^n} = 195.952381.$

As there is no $x$ and $n$ in the allowed intervals such that the last equation holds true, there must be no solutions.

Just use excel Remember one fact: [32x] respondes one to small variations of x, namely, step of 1/32 if de decimal portion of x is between 0 and 1/32, the expression becames: 63y = 12345, where y is the integer part of x; result is decimal - no solution Now, iterate in steps of 1/32. Use Excell Each cell has the formula: =FLOOR(F$3*$E4;1)

It could be done with expressions but it's to boring with no real value added

Step 1/32   1   2   4   8   16  32

1 0,03115 0 0 0 0 0 0 63 195,952381 2 0,0624 0 0 0 0 0 1 64 192,890625 3 0,09365 0 0 0 0 1 2 66 187,0454545 4 0,1249 0 0 0 0 1 3 67 184,2537313 5 0,15615 0 0 0 1 2 4 70 176,3571429 6 0,1874 0 0 0 1 2 5 71 173,8732394 7 0,21865 0 0 0 1 3 6 73 169,109589 8 0,2499 0 0 0 1 3 7 74 166,8243243 9 0,28115 0 0 1 2 4 8 78 158,2692308 10 0,3124 0 0 1 2 4 9 79 156,2658228 11 0,34365 0 0 1 2 5 10 81 152,4074074 12 0,3749 0 0 1 2 5 11 82 150,5487805 13 0,40615 0 0 1 3 6 12 85 145,2352941 14 0,4374 0 0 1 3 6 13 86 143,5465116 15 0,46865 0 0 1 3 7 14 88 140,2840909 16 0,4999 0 0 1 3 7 15 89 138,7078652 17 0,53115 0 1 2 4 8 16 94 131,3297872 18 0,5624 0 1 2 4 8 17 95 129,9473684 19 0,59365 0 1 2 4 9 18 97 127,2680412 20 0,6249 0 1 2 4 9 19 98 125,9693878 21 0,65615 0 1 2 5 10 20 101 122,2277228 22 0,6874 0 1 2 5 10 21 102 121,0294118 23 0,71865 0 1 2 5 11 22 104 118,7019231 24 0,7499 0 1 2 5 11 23 105 117,5714286 25 0,78115 0 1 3 6 12 24 109 113,2568807 26 0,8124 0 1 3 6 12 25 110 112,2272727 27 0,84365 0 1 3 6 13 26 112 110,2232143 28 0,8749 0 1 3 6 13 27 113 109,2477876 29 0,90615 0 1 3 7 14 28 116 106,4224138 30 0,9374 0 1 3 7 14 29 117 105,5128205 31 0,96865 0 1 3 7 15 30 119 103,7394958 32 0,9999 0 1 3 7 15 31 120 102,875

let x be an integer ... so [x]=x,hence x+2x+4x+8x+16x+32x = 12345 ..... solving this equation we get x=195.95 which is not an integer . so it contradicts above assumption hence no solution