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# so confused

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Let $$f(x) = x - \lfloor \sqrt{x} \rfloor^2$$. What is $$f(101) + f(102) + f(103) + \cdots + f(110)?$$

Jun 15, 2020

#1
+10080
+1

Let  $$f(x) = x - \lfloor \sqrt{x} \rfloor^2$$ What is $$f(101) + f(102) + f(103) + \cdots + f(110)?$$

Hello Guest!

$$f(x) = x - \lfloor \sqrt{x} \rfloor^2$$

$$f(101) + f(102) + f(103) + \cdots + f(110)=0$$

!

Jun 16, 2020
#2
0

i tried that but that was incorrect

Guest Jun 16, 2020
#3
+179
-1

Oh god

Varxaax  Jun 16, 2020
#4
0

wdym oh god

Guest Jun 16, 2020
#5
+110715
+1

It is not going to be 0 because the floor of sqrtx is usually smaller than the sqrtx.

Have you tried to work this out yourself guest?

Show us what you have tried.

Hint: there is only 10 functions there. Just do them one by one.  Easy Peazy.   (I expect)

Jun 16, 2020
#6
+10080
+1

$$f(x) = x - \lfloor \sqrt{x} \rfloor^2\\ f(101) = 101 - \lfloor \sqrt{101} \rfloor^2=0\\ f(102) = 102 - \lfloor \sqrt{102} \rfloor^2=0\\ What\ is\ wrong?\\ Greetings$$

!

asinus  Jun 16, 2020
#7
+110715
+2

Perhaps you are unfamiliar with the floor function asinus.

$$f(x) = x - \lfloor \sqrt{x} \rfloor^2\\ f(101)=101-\lfloor \sqrt{101} \rfloor^2=101-10^2=101-100=1 \\ f(102)=102-100=2\\ f(103)=3\\ ...\\ f(110)=10\\$$

So the sum is 1+2+3+...+10 = 1+10+2+9+3+8+4+7+5+6 = 11*5 = 55

I have not checked my answer carefully.

Jun 16, 2020
edited by Melody  Jun 16, 2020
#8
+10080
+1

Exactly, I find out for the first time about the existence of the floor function.

Thank you, Melody

!

asinus  Jun 16, 2020