Alphabets and Numbers? : Another Grandpa Mystery

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
3
down vote

favorite
1












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!”










share|improve this question



















  • 1




    Is the ** a formating error?
    – PotatoLatte
    1 hour ago










  • @PotatoLatte Maybe, maybe not :P
    – user477343
    50 mins ago










  • Well, since it got fixed, no?
    – PotatoLatte
    41 mins ago










  • It probably does not have to do with the numbers themselves, as they use the words instead of the numbers
    – PotatoLatte
    40 mins ago










  • 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
    40 mins ago














up vote
3
down vote

favorite
1












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!”










share|improve this question



















  • 1




    Is the ** a formating error?
    – PotatoLatte
    1 hour ago










  • @PotatoLatte Maybe, maybe not :P
    – user477343
    50 mins ago










  • Well, since it got fixed, no?
    – PotatoLatte
    41 mins ago










  • It probably does not have to do with the numbers themselves, as they use the words instead of the numbers
    – PotatoLatte
    40 mins ago










  • 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
    40 mins ago












up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





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!”










share|improve this question















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!”







riddle lateral-thinking






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago









El-Guest

15.1k3272




15.1k3272










asked 1 hour ago









DEEM

4,1451179




4,1451179







  • 1




    Is the ** a formating error?
    – PotatoLatte
    1 hour ago










  • @PotatoLatte Maybe, maybe not :P
    – user477343
    50 mins ago










  • Well, since it got fixed, no?
    – PotatoLatte
    41 mins ago










  • It probably does not have to do with the numbers themselves, as they use the words instead of the numbers
    – PotatoLatte
    40 mins ago










  • 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
    40 mins ago












  • 1




    Is the ** a formating error?
    – PotatoLatte
    1 hour ago










  • @PotatoLatte Maybe, maybe not :P
    – user477343
    50 mins ago










  • Well, since it got fixed, no?
    – PotatoLatte
    41 mins ago










  • It probably does not have to do with the numbers themselves, as they use the words instead of the numbers
    – PotatoLatte
    40 mins ago










  • 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
    40 mins ago







1




1




Is the ** a formating error?
– PotatoLatte
1 hour ago




Is the ** a formating error?
– PotatoLatte
1 hour ago












@PotatoLatte Maybe, maybe not :P
– user477343
50 mins ago




@PotatoLatte Maybe, maybe not :P
– user477343
50 mins ago












Well, since it got fixed, no?
– PotatoLatte
41 mins ago




Well, since it got fixed, no?
– PotatoLatte
41 mins ago












It probably does not have to do with the numbers themselves, as they use the words instead of the numbers
– PotatoLatte
40 mins ago




It probably does not have to do with the numbers themselves, as they use the words instead of the numbers
– PotatoLatte
40 mins ago












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
40 mins ago




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
40 mins ago










2 Answers
2






active

oldest

votes

















up vote
4
down vote













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.







share|improve this answer
















  • 1




    Damn, I reached my daily voting limit (DVL).
    – user477343
    22 mins ago


















up vote
3
down vote













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 $x = mboxSeven+L = 7+50=57$, $f(x) = (57+1) pmod 2 = 0 = rm Zero$.
For $x = textZero-C = 0-100=-100$, $f(x) = (-100+1) pmod 2 = 1 = rm One$.
For $x= mathrmThree + I = 3+1 = 4$, $f(x) = (4+1) pmod 2 = 1 = rm One$.




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 $xgeq 0$ and 0 for $x<0$.




Then




For $x = mboxSeven+L = 7+50=57$, $f(x) = H(-57) = 0 = rm Zero$.
For $x = textZero-C = 0-100=-100$, $f(x) = H(-(-100)) = 1 = rm One$.
For $x= mathrmThree + I = 3+1 = 4$, $f(x) = H(-4) = 0 = rm Zero$.







share|improve this answer


















  • 1




    Please note that there is no MATH tag on this @El-Guest
    – DEEM
    54 mins ago










  • 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 $rm underlinethis~font$ 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
    44 mins ago







  • 1




    Yes, but I figured there was some lateral thinking involved in deriving this answer! Back to the drawing board!
    – El-Guest
    42 mins ago






  • 1




    Thank you, @user477343! Much appreciated!
    – El-Guest
    41 mins ago










  • No problemo! $(+1)$ :D
    – user477343
    41 mins ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote













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.







share|improve this answer
















  • 1




    Damn, I reached my daily voting limit (DVL).
    – user477343
    22 mins ago















up vote
4
down vote













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.







share|improve this answer
















  • 1




    Damn, I reached my daily voting limit (DVL).
    – user477343
    22 mins ago













up vote
4
down vote










up vote
4
down vote









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.







share|improve this answer












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.








share|improve this answer












share|improve this answer



share|improve this answer










answered 25 mins ago









El-Guest

15.1k3272




15.1k3272







  • 1




    Damn, I reached my daily voting limit (DVL).
    – user477343
    22 mins ago













  • 1




    Damn, I reached my daily voting limit (DVL).
    – user477343
    22 mins ago








1




1




Damn, I reached my daily voting limit (DVL).
– user477343
22 mins ago





Damn, I reached my daily voting limit (DVL).
– user477343
22 mins ago











up vote
3
down vote













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 $x = mboxSeven+L = 7+50=57$, $f(x) = (57+1) pmod 2 = 0 = rm Zero$.
For $x = textZero-C = 0-100=-100$, $f(x) = (-100+1) pmod 2 = 1 = rm One$.
For $x= mathrmThree + I = 3+1 = 4$, $f(x) = (4+1) pmod 2 = 1 = rm One$.




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 $xgeq 0$ and 0 for $x<0$.




Then




For $x = mboxSeven+L = 7+50=57$, $f(x) = H(-57) = 0 = rm Zero$.
For $x = textZero-C = 0-100=-100$, $f(x) = H(-(-100)) = 1 = rm One$.
For $x= mathrmThree + I = 3+1 = 4$, $f(x) = H(-4) = 0 = rm Zero$.







share|improve this answer


















  • 1




    Please note that there is no MATH tag on this @El-Guest
    – DEEM
    54 mins ago










  • 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 $rm underlinethis~font$ 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
    44 mins ago







  • 1




    Yes, but I figured there was some lateral thinking involved in deriving this answer! Back to the drawing board!
    – El-Guest
    42 mins ago






  • 1




    Thank you, @user477343! Much appreciated!
    – El-Guest
    41 mins ago










  • No problemo! $(+1)$ :D
    – user477343
    41 mins ago















up vote
3
down vote













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 $x = mboxSeven+L = 7+50=57$, $f(x) = (57+1) pmod 2 = 0 = rm Zero$.
For $x = textZero-C = 0-100=-100$, $f(x) = (-100+1) pmod 2 = 1 = rm One$.
For $x= mathrmThree + I = 3+1 = 4$, $f(x) = (4+1) pmod 2 = 1 = rm One$.




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 $xgeq 0$ and 0 for $x<0$.




Then




For $x = mboxSeven+L = 7+50=57$, $f(x) = H(-57) = 0 = rm Zero$.
For $x = textZero-C = 0-100=-100$, $f(x) = H(-(-100)) = 1 = rm One$.
For $x= mathrmThree + I = 3+1 = 4$, $f(x) = H(-4) = 0 = rm Zero$.







share|improve this answer


















  • 1




    Please note that there is no MATH tag on this @El-Guest
    – DEEM
    54 mins ago










  • 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 $rm underlinethis~font$ 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
    44 mins ago







  • 1




    Yes, but I figured there was some lateral thinking involved in deriving this answer! Back to the drawing board!
    – El-Guest
    42 mins ago






  • 1




    Thank you, @user477343! Much appreciated!
    – El-Guest
    41 mins ago










  • No problemo! $(+1)$ :D
    – user477343
    41 mins ago













up vote
3
down vote










up vote
3
down vote









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 $x = mboxSeven+L = 7+50=57$, $f(x) = (57+1) pmod 2 = 0 = rm Zero$.
For $x = textZero-C = 0-100=-100$, $f(x) = (-100+1) pmod 2 = 1 = rm One$.
For $x= mathrmThree + I = 3+1 = 4$, $f(x) = (4+1) pmod 2 = 1 = rm One$.




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 $xgeq 0$ and 0 for $x<0$.




Then




For $x = mboxSeven+L = 7+50=57$, $f(x) = H(-57) = 0 = rm Zero$.
For $x = textZero-C = 0-100=-100$, $f(x) = H(-(-100)) = 1 = rm One$.
For $x= mathrmThree + I = 3+1 = 4$, $f(x) = H(-4) = 0 = rm Zero$.







share|improve this answer














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 $x = mboxSeven+L = 7+50=57$, $f(x) = (57+1) pmod 2 = 0 = rm Zero$.
For $x = textZero-C = 0-100=-100$, $f(x) = (-100+1) pmod 2 = 1 = rm One$.
For $x= mathrmThree + I = 3+1 = 4$, $f(x) = (4+1) pmod 2 = 1 = rm One$.




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 $xgeq 0$ and 0 for $x<0$.




Then




For $x = mboxSeven+L = 7+50=57$, $f(x) = H(-57) = 0 = rm Zero$.
For $x = textZero-C = 0-100=-100$, $f(x) = H(-(-100)) = 1 = rm One$.
For $x= mathrmThree + I = 3+1 = 4$, $f(x) = H(-4) = 0 = rm Zero$.








share|improve this answer














share|improve this answer



share|improve this answer








edited 46 mins ago









user477343

2,5281748




2,5281748










answered 1 hour ago









El-Guest

15.1k3272




15.1k3272







  • 1




    Please note that there is no MATH tag on this @El-Guest
    – DEEM
    54 mins ago










  • 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 $rm underlinethis~font$ 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
    44 mins ago







  • 1




    Yes, but I figured there was some lateral thinking involved in deriving this answer! Back to the drawing board!
    – El-Guest
    42 mins ago






  • 1




    Thank you, @user477343! Much appreciated!
    – El-Guest
    41 mins ago










  • No problemo! $(+1)$ :D
    – user477343
    41 mins ago













  • 1




    Please note that there is no MATH tag on this @El-Guest
    – DEEM
    54 mins ago










  • 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 $rm underlinethis~font$ 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
    44 mins ago







  • 1




    Yes, but I figured there was some lateral thinking involved in deriving this answer! Back to the drawing board!
    – El-Guest
    42 mins ago






  • 1




    Thank you, @user477343! Much appreciated!
    – El-Guest
    41 mins ago










  • No problemo! $(+1)$ :D
    – user477343
    41 mins ago








1




1




Please note that there is no MATH tag on this @El-Guest
– DEEM
54 mins ago




Please note that there is no MATH tag on this @El-Guest
– DEEM
54 mins ago












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 $rm underlinethis~font$ 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
44 mins ago





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 $rm underlinethis~font$ 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
44 mins ago





1




1




Yes, but I figured there was some lateral thinking involved in deriving this answer! Back to the drawing board!
– El-Guest
42 mins ago




Yes, but I figured there was some lateral thinking involved in deriving this answer! Back to the drawing board!
– El-Guest
42 mins ago




1




1




Thank you, @user477343! Much appreciated!
– El-Guest
41 mins ago




Thank you, @user477343! Much appreciated!
– El-Guest
41 mins ago












No problemo! $(+1)$ :D
– user477343
41 mins ago





No problemo! $(+1)$ :D
– user477343
41 mins ago


















 

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