# Alphabets and Numbers? : Another Grandpa Mystery

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!”

• Is the ** a formating error? – PotatoLatte Oct 12 '18 at 12:35
• @PotatoLatte Maybe, maybe not :P – Mr Pie Oct 12 '18 at 13:06
• Well, since it got fixed, no? – PotatoLatte Oct 12 '18 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 '18 at 13:16
• H - Four + Five = ? is another nice one. – Keeta - reinstate Monica Oct 12 '18 at 19:09

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.

• Damn, I reached my daily voting limit (DVL). – Mr Pie Oct 12 '18 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 $$x= \mathrm{Three} + I = 3+1 = 4$$, $$f(x) = H(-4) = 0 = \rm Zero$$.

• Please note that there is no MATH tag on this @El-Guest – DrD Oct 12 '18 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? :) – Mr Pie Oct 12 '18 at 13:12
• Yes, but I figured there was some lateral thinking involved in deriving this answer! Back to the drawing board! – El-Guest Oct 12 '18 at 13:14
• Thank you, @user477343! Much appreciated! – El-Guest Oct 12 '18 at 13:15
• No problemo! $(+1)$ :D – Mr Pie Oct 12 '18 at 13:15