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!”
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Sign up to join this communityNow 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!”
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.
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 $x= \mathrm{Three} + I = 3+1 = 4$, $f(x) = H(-4) = 0 = \rm Zero$.
\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? :)
$\endgroup$