Floor Sums

Let $x = \lfloor x \rfloor + \{ x \}$ , where $\{x\}$ is the decimal part of $x$ . Then $\begin{aligned} 331 & = \left \lfloor x +\frac {11}{100} \right \rfloor + \left \lfloor x + \frac {12}{100} \right \rfloor + \ldots + \left \lfloor x + \frac {90} {100} \right \rfloor \\ & = \left \lfloor \left \lfloor x \right \rfloor + \{ x \} +\frac {11}{100} \right \rfloor + \left \lfloor \left \lfloor x \right \rfloor + \{ x \} + \frac {12}{100} \right \rfloor + \ldots + \left \lfloor \left \lfloor x \right \rfloor + \{ x \} + \frac {90} {100} \right \rfloor \\ & = 80 \lfloor x \rfloor + \left \lfloor \{ x \} +\frac {11}{100} \right \rfloor + \left \lfloor \{ x \} + \frac {12}{100} \right \rfloor + \ldots + \left \lfloor \{ x \} + \frac {90} {100} \right \rfloor \\ \end{aligned}$

For $11 \leq i \leq 90$ , $\left \lfloor \{ x \} + \frac {i} {100} \right \rfloor = 0$ or $1$ and is increasing in $i$ . The minimum value of $\lfloor x \rfloor$ occurs when they are all equal to $1$ . Thus $\lfloor x \rfloor \geq \frac{331-80}{80} = \frac{251}{80}$ . The maximum value of $\lfloor x \rfloor$ occurs when they are all equal to $0$ . Thus $\lfloor x \rfloor \leq \frac{331}{80}$ . Therefore $\frac{251}{80} \leq \lfloor x \rfloor \leq \frac {331}{80} \Rightarrow \lfloor x \rfloor = 4$ .

Thus the rest of the terms must sum to $331-80\cdot 4 = 11$ which implies that the last $11$ terms must be $1$ and the rest of terms must be $0$ . Since $90-11+1 = 80$ , we get that $\left \lfloor \{ x \} + \frac {80}{100} \right \rfloor = 1$ and that $\left \lfloor \{ x \} + \frac { 79}{100} \right \rfloor = 0$ . This gives us $\frac {20} {100} \leq \{ x \} < \frac {21}{100}$ .

Therefore $\frac {420}{100} \leq x < \frac {421}{100}$ $\Rightarrow \lfloor 100 x \rfloor = 420$ .

Hermite's Identity state that summation from k=0 to k=n-1 of [x + k/n] = [nx] and let n=100

So that, [100x] - summation from k=0 to k=10 of [x + k/100] - summation from k=91 to k=99 of [x+k/100] It obvious to see that, summation from k=91 to k=99 of [x+k/100] = 9 5=45 and summation from k=0 to k=10 of [x + k/100] = 11 4=44

so , [100x] = 331+44+45 = 420

Another approach: There are 80 terms, and notice that 331 = 4*80 + 11, which means the last 11 terms are 5 (n=80 to n=90).

Then, [x + 80/100] = 5 which imply x >= 4.20 and [x+79/100] = 4 which imply x<4.21

Hence, [100x] = 420

Here [x] is a floor function

Here, 331 is expressed as the sum of 80 integers. Let's first find out what the integers are. Since $(x+0.90)-(x+0.11)=0.79$ , $\lfloor x+0.90\rfloor -\lfloor x+0.11\rfloor$ will not exceed 1. Therefore, the 80 integers are either $n$ or $n+1$ , where $n$ is an integer satisfying $80 n \leq 331 \leq 80 (n+1)$ . Solving this inequality, we get $n=4$ .

Now, let's identify the number of fours and fives. If there are $x$ fours and $(80-x)$ fives, 331 can be expressed as $4x+5(80-x)=331$ , which yields $x=69$ . So there are 69 fours and 11 fives in the equation. That means the eleventh-to-last number \lfloor x+0.80 \rfloor =5), and the twelfth-to-last number $\lfloor x+0.79 \rfloor =4$ . Hence, $5 \leq x+0.80 < 6$ and $4 \leq x+0.79 < 5$ . This tells us that $4.20 \leq x < 4.21$ , $420 ≤ 100x < 421$ , \lfloor 100x \rfloor =420).

Let $n=\lfloor x+\frac{11}{100}\rfloor \leq \lfloor x+\frac{81}{100}\rfloor \leq \lfloor x+\frac{111}{100}\rfloor=n+1$ . Let there be $a$ terms in the expression be equal to $n$ . Therefore $331=an+(80-a)(n+1)=80n+80-a$ . Since $0 \leq 80-a \leq 79, n=\lfloor \frac{331}{80}\rfloor=4$ . Then solving the linear equation $331=400-a$ gives $a=69$ , which means $\lfloor x+\frac{79}{100}\rfloor=4, \lfloor x+\frac{80}{100}\rfloor=5 \Rightarrow x=4.2 \Rightarrow \lfloor100x\rfloor=420$ .

There are 80 terms in the equation.So, 331 divide by 80 is 4.1375. It indicates than some of the terms are 5. Since 331=4(69)+5(11), the 70th terms will exactly equal to 5 when in 2 decimal places and the 69th term will be the nearest to 5 in 2 decimal places, which is 4.99. So x=4.20 and the floor function of 100x is 420.

[] denotes floor as there is no better symbol in the keyboard if we consider [x+11/100] as n, [x+90/100] can be n or n+1. Because x+90/100 = (x+11/100)+79/100 < (n+1)+79/100<n+2, resulting in n+2>x+90/100>x+11/100>=n. By definition of floor, it's obvious that [x+90/100] is n or n+1. We can see that it's similar for those 80 in the addition. any of those floor, other than [x+11/100] can be either n or n+1. If we consider that there are m of those floors that values n+1, we can see that n (80-m) + m(n+1) = 331 80n+m = 331 m congr 11 mod 80 as m can only range from 0 to 79, m is definitely 11. 80n+11=331 n=4 we now know that 11 of those floors result in 5. It's definitely the 11 largest, because if any floor before them equals 5, the 11 largest will equal 5 as well, resulting in at least 12 floors valuing 5. So, taking the smallest of the 11 largest, we get [x+80/100] = 5 meaning 5<= (x+80/100)<600 500 <= 100x+80 <600 420<=100x<520 (i)

but if we take the floor beside it, we get [x+79/100]=4 4<=x+79/100<5 400<=100x+79<500 421<=100x<421 (ii)

from (i) and (ii), we get that 420<=100x<421 By definition of floor, [100x]=420

there are total 80 floors ... and then 80 X approx to 331 then X closes to 4.13.. BUT 80X approximates to 320.. here, we need to have 11 more to get 331.. now here comes the importance of real part of X.. total floor numbers from floor{X+80/100} to floor{X+90/100} are equal to 11.. so we can chose the real part of X as 4.20... adding 80/100 to 4.20 gives us floor{5.00} =5.00 and this continues till floor {X+90/100}.. eg:

floor{X+81/100}= floor{4.20+0.81}=floor{5.01}=5.00

floor{X+82/100}= floor{4.20+0.82}=floor{5.02}=5.00

floor{X+83/100}= floor{4.20+0.83}=floor{5.03}=5.00

floor{X+84/100}= floor{4.20+0.84}=floor{5.04}=5.00

and so on floor{X+90/100}= floor{4.20+0.90}=floor{5.10}=5.00

so numbers from floor{X+80/100} to floor{X+90/100} gives u: 5+5+5+... 11 times that adds up to 55..

and numbers from floor{X+11/100} to floor{X+79/100} gives us 4+4+.... 69 times : 4*69=276

now adding up 276+55=331

now X is 4.20

floor{100 X}= floor{100 4.20}= floor {420}=420

Lets first look at this equation: $x+x+x+\cdots_{\text{90-10=80 times}}=331$ So we get $x=4.\text{something}$ This means that one of the values of the individual terms ( $[.]$ ) are $4$ . But it cannot be so because the RHS is even. So this means that somewhere, the value changes to $5$ . It can be easily seen why it cannot change to $6$ . So we have equations: $4a+5b=331$ and $a+b=80$ . Solving both of them, we get $a=69$ , $b=11$ . So the value of the terms remains $4$ till the $69$ th term. And it will be $\lfloor x+\frac{79}{100}\rfloor$ . Now we want its value to be $4.99$ , so we get that $\boxed{x=4.20}$ So the required answer is easily, $420$ .

Assume that k=100*x is a positive integer. By iterate over integers, we can find k=420. => x=4.2, 100x=420.