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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1$\begingroup$ Is the ** a formating error? $\endgroup$ – PotatoLatte Oct 12 '18 at 12:35
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$\begingroup$ @PotatoLatte Maybe, maybe not :P $\endgroup$ – Mr Pie Oct 12 '18 at 13:06
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$\begingroup$ Well, since it got fixed, no? $\endgroup$ – PotatoLatte Oct 12 '18 at 13:15
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$\begingroup$ It probably does not have to do with the numbers themselves, as they use the words instead of the numbers $\endgroup$ – PotatoLatte Oct 12 '18 at 13:16
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1$\begingroup$ H - Four + Five = ? is another nice one. $\endgroup$ – Keeta - reinstate Monica Oct 12 '18 at 19:09
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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.
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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 $x= \mathrm{Three} + I = 3+1 = 4$, $f(x) = H(-4) = 0 = \rm Zero$.
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1$\begingroup$ Please note that there is no MATH tag on this @El-Guest $\endgroup$ – DEEM Oct 12 '18 at 13:02
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$\begingroup$ I have edited this post by carrying out the following actions: changed $mod2$ to $\pmod 2$ via the command
\pmod 2and I also changed $this~font$ to $\rm \underline{this~font}$ with multiple commands like\mboxand\textetc (see which one you prefer). (The underline is not part of the actual font, but provides an emphasis.) Do you approve of this edit? :) $\endgroup$ – Mr Pie Oct 12 '18 at 13:12 -
1$\begingroup$ Yes, but I figured there was some lateral thinking involved in deriving this answer! Back to the drawing board! $\endgroup$ – El-Guest Oct 12 '18 at 13:14
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