What is this symbol??

[Guest]

Hello, I’m kinda new to this form and found it in Google while trying to find an answer to this question... so I thought I’d post it hear and give you people a crack at it

This is all the relevant information (exactly as I received it):

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Given that:

5 # 3 = 4

2 # 8 = 2

6 # 3 = 3


What is:

1 # 7 = ?


What does the # represent in the problem?
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I honestly don’t know if there is an answer, I got this from my physics teacher in an e-mail and it had all the maths and physics teachers e-mails on it from my school and from the list it has been to maths teachers from other schools as well and none have solved it (as of 26 of May, 2006).

The only contribution I seemed to have made (at my school at least) is that all the numbers in the equation add up to 12 (e.g. 5 + 3 + 4 = 12) so my assumption (I’m making this clear that this is a guess and might be solidly coincidence) is that 1 # 7 = 4, but I have no proof and that ultimately isn’t what the question is asking anyway.

I’m stuck, I haven’t found any other links with the numbers or any other help on the internet, plus I don’t know what type of maths it is, I hope it gives you guys a challenge (if you haven’t already seen it, lol) and that an answer is found!

Thanks for any help!!

 

[soroban]

Hello, Boomar!

Someone is bound to point out that there are zillions of right answers. **
\(\displaystyle \;\;\)But I'll assume that there is a pattern involved.

It's an exercise in Pattern Recognition.
\(\displaystyle \;\;\)And I think you've nailed the answer!

Given that: \(\displaystyle \,5\,#\,3\:=\:4,\;\;2\,#\,8\:=\:2,\;\;6\,#\,3\:=\:3\)

What is: \(\displaystyle \,1\,#\,7\) ?

What does the \(\displaystyle #\) represent in the problem?

The only contribution I made is that all the numbers in the equation add up to 12. . . . That's it!

It would seem that: \(\displaystyle \,a\,#\,b\) means: "Add \(\displaystyle a\) and \(\displaystyle b\), and subtract from \(\displaystyle 12.\)"

In general: \(\displaystyle \,a\,#\,b\;=\;12\,-\,(a\,+\,b)\)

So your guess is correct: \(\displaystyle \,1\,#\,7\:=\:12\,-\,(1\,+\,7)\;=\;4\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**
To give you a taste of the zillions . . .


\(\displaystyle \text{A solution could be: }\L\,a\,#\,b\;=\;\frac{474\,-\,5a^2\,-\,43b}{55}\)

This says: "Start with 474, subtract 5 times the square of a,
\(\displaystyle \;\;\)subtract 43 times b, then divide by 55."

This works for the three given equations.

Then: \(\displaystyle \,1\,#\,7\;=\;\frac{474\,-\,5\cdot1^2\,-\,43\cdot7}{55} \;= \;\frac{168}{55}\)


\(\displaystyle \text{Another could be: }\L\,a\,#\,b\;=\;\frac{108\,-\,11a\,-\,b^2}{11}\)

It says: "Start with 108, subtract 11 times a,
\(\displaystyle \;\;\)subtract the square of b, then divide by \(\displaystyle 11.\)

This too work for the three given equations.

Then: \(\displaystyle \,1\,#\,7\;=\;\frac{108\,-\,11\cdot1\,-\,7^2}{11}\;=\;\frac{48}{11}\)


And we'd better get full credit for these answers . . . they're correct!

 

[Guest]

Wow, thanks a lot!
I’m normally the one that picks pattern formulas like that out to others, lol. I understood everything, it just didn’t click that how I ended up getting 12 was the answer, I just never put it as an algebraic formula like that, and it passed all my teachers as well (and math’s coordinators)... gosh I will admit that the answer seems incredibly simple and I cant believe it slipped by them and myself.
Again, that’s a lot, and for your response speed as well!