Now we all know how crazy Grandpa is when it comes to Math. He says to me:

"Use your imagination and tell me,

If
Seven + L = Zero
And
Zero - C = One
Then
Three + I = ?

I have no idea!”

  • 1
    Is the ** a formating error? – PotatoLatte Oct 12 at 12:35
  • @PotatoLatte Maybe, maybe not :P – user477343 Oct 12 at 13:06
  • Well, since it got fixed, no? – PotatoLatte Oct 12 at 13:15
  • It probably does not have to do with the numbers themselves, as they use the words instead of the numbers – PotatoLatte Oct 12 at 13:16
  • It was fixed by me because I assumed this was the intention. If DEEM wishes to change it, they may roll it back. – El-Guest Oct 12 at 13:16
up vote 32 down vote accepted

Based on DEEM’s hint, the answer is

Eight.

I think this is because

When you look at the digital number for 7, and add an L to the left of it, you get L7 which looks like a digital Zero. When you look at a digital Zero and remove a C from the left of it, you get a digital One. When you take a digital 3 and add an I to the left of it, you get a digital Eight.

  • 1
    Damn, I reached my daily voting limit (DVL). – user477343 Oct 12 at 13:34

I think it should be

One, also.

The function is

$f(x) = (x+1) \pmod 2$, where $x$ is the result of the addition or subtraction when you convert the letters to Roman numerals.

So:

For $x = \mbox{Seven}+L = 7+50=57$, $f(x) = (57+1) \pmod 2 = 0 = \rm Zero$.
For $x = \text{Zero}-C = 0-100=-100$, $f(x) = (-100+1) \pmod 2 = 1 = \rm One$.
For $x= \mathrm{Three} + I = 3+1 = 4$, $f(x) = (4+1) \pmod 2 = 1 = \rm One$.

Of course, the answer could also be

Zero.

If the function was

$f(x) = H(-x)$, where $H$ is the Heaviside function, equal to 1 for $x\geq 0$ and 0 for $x<0$.

Then

For $x = \mbox{Seven}+L = 7+50=57$, $f(x) = H(-57) = 0 = \rm Zero$.
For $x = \text{Zero}-C = 0-100=-100$, $f(x) = H(-(-100)) = 1 = \rm One$.
For $x= \mathrm{Three} + I = 3+1 = 4$, $f(x) = H(-4) = 0 = \rm Zero$.

  • 1
    Please note that there is no MATH tag on this @El-Guest – DEEM Oct 12 at 13:02
  • I have edited this post by carrying out the following actions: changed $mod2$ to $\pmod 2$ via the command \pmod 2 and I also changed $this~font$ to $\rm \underline{this~font}$ with multiple commands like \mbox and \text etc (see which one you prefer). (The underline is not part of the actual font, but provides an emphasis.) Do you approve of this edit? :) – user477343 Oct 12 at 13:12
  • 1
    Yes, but I figured there was some lateral thinking involved in deriving this answer! Back to the drawing board! – El-Guest Oct 12 at 13:14
  • 1
    Thank you, @user477343! Much appreciated! – El-Guest Oct 12 at 13:15
  • No problemo! $(+1)$ :D – user477343 Oct 12 at 13:15

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