## Callisto Group Title In the domain [0, 100], how many solutions are there for the equation $x^2- \lfloor x \rfloor x = 81.25$ How to start? one year ago one year ago

1. joemath314159 Group Title

Instead of thinking of it as one function defined on [1,100], maybe think of it as 100 different function defined on the intervals [n,n+1] as n goes from 0 to 99.

2. joemath314159 Group Title

That way you can minimize the frustration the floor function brings to the table.

3. joemath314159 Group Title

er, i guess the interval should really be [n,n+1)

4. joemath314159 Group Title

Then on each interval, the question is really:$x^2+nx-81.25=0$where x ranges from [n,n+1). Not sure if this leads to a solution, but it might be a better way of thinking of the problem.

5. Callisto Group Title

Domain = [0, 100] :S

6. joemath314159 Group Title

wait, the function$x^2+\lfloor x \rfloor x$is strictly increasing, isnt it?

7. joemath314159 Group Title

then it can have at most one solution.

8. Callisto Group Title

That means it has 0 :'(

9. joemath314159 Group Title

strictly increasing on the interval [0,100] anyways.

10. joemath314159 Group Title

well, im not 100% sure on the strictly increasing part, im a little fuzzy on what happens when you jump from one unit interval to the other.

11. joemath314159 Group Title

no, its strictly increasing for sure.

12. Callisto Group Title

Sorry!!! Wait... a mistake in the question posted here!!

13. joemath314159 Group Title

ah ok, that makes more sense.

14. Callisto Group Title

I'm sorry!!!! :(

15. joemath314159 Group Title

if you look at the function:$x^2-\lfloor x \rfloor x$restricted to an interval {n,n+1) (open on the right), note that the values start at 0 when x = n, and as x approaches n+1, the values also approach n+1:|dw:1357103977725:dw|

16. joemath314159 Group Title

So the equation $x^2-\lfloor x \rfloor x=81.25$only has a solution when $x\le 81$the interval {81,82) is going to be the first time the parabola piece gets as large as 81.25.

17. joemath314159 Group Title

less than >.> i meant greater than lol.

18. joemath314159 Group Title

$x\ge 81.25$

19. joemath314159 Group Title

>.<$x\ge 81$

20. joemath314159 Group Title

So there are 19 solutions, one in each of the intervals [81, 82), [82,83),....., [99,100)

21. Callisto Group Title

Thanks! I've never thought about this way to look at the problem.. :(

22. joemath314159 Group Title

The idea is that the floor function on large intervals in a pain, but on unit intervals, its just a the constant function. an interesting problem indeed :)

23. Callisto Group Title

I'll remember it! Thanks again!!

24. shubhamsrg Group Title

x-[x] = {x} right.. ? {x} = fractional part of x so if we simplify a little, we see we have x*{x} = 81.25 does this help ?

25. shubhamsrg Group Title

x= 81.25 /{x} x can go max to 100 so fractional part can be minimum 0.8125 and max 0.999999...

26. shubhamsrg Group Title

just substitute in new values of {x} to get new values of x so according to me,,infinite solutions..

27. shubhamsrg Group Title

ohh wait,,did i make a mistake in logic,,i am getting a feeling,,!??

28. shubhamsrg Group Title

@joemath314159 @mukushla @calculusfunctions

29. joemath314159 Group Title

if you look at the function:$\frac{81.25}{\{x\}}$ since the fractional part of x can only go from 0 to 1, the only time that will have a solution is when x is bigger than 81.

30. joemath314159 Group Title

if you look at domain, say, [79, 80), this equation:$x=\frac{81.25}{\{x\}}$wont have a solution. since the smallest$\frac{81.25}{\{x\}}$can be is 81.25

31. shubhamsrg Group Title

maybe this is where am flawed: since we are dealing with fractional part separately, x , cannot have a fractional part..!! like for example if we set {x} = 0.9, x = 90.2777.. hence contradiction --> 2 different fractional parts..

32. shubhamsrg Group Title

i see,,only solutions will be when x is an integer..so that the contradiction with fractional part doesnt come into play and since x has to be greater than 81.25 as @joemath314159 rightly pointed, x can take values 82,83...100 -->19 solutions..

33. shubhamsrg Group Title

i mean x can not take those values,, but there'll be a solution each in [82,83) and so on.. so yes,,19 solutions..

34. mukushla Group Title

problem says how many solutions so its done...and this is just for fun :) we can find exact solutions:$[x]+\{x\}=\frac{81.25}{\{x\}}$$[x]=\frac{81.25}{\{x\}}-\{x\}$now we know that $[x]=\{81,82,...,99\}$so for example for the solution that lies on the interval [81,82) we have$81=\frac{81.25}{\{x\}}-\{x\}$this is a quadratic for $$\{x\}$$

35. shubhamsrg Group Title

yep..i understand better now..thank you..