Here's a maths problem for you folks:
Complete the following equations using mathematical operators (but no numbers(*)):
| Code: |
0 0 0 = 6
1 1 1 = 6
2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6
10 10 10 = 6
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(*) For example, you may use the square root operator, but not the square operator (uses number 2) or the cube root operator (uses number 3).
You'll find that several equations are easy to complete, but others are more stubborn.
When you've done the easy ones, concentrate on on the first.
Good luck! Are we also to presume that all numbers are in decimal format rather than using different bases?
Hi Bikerman
Does this mean that from now on all mathematical texts must start by declaring what number base is used? 
I presume (but maybe I'm wrong) that this tongue-in-cheek remark is due to having read my post on the rule for numbers divisible by 9...
Well, since you ask we are using base 10 
- I mean 9+1. The question was perfectly serious - not at all tongue in cheek. If you use different number bases within each expression, for example:
10 +10 +10 = 6 is easy (10 base 2, and 6 base 7 or greater).....
However, I'll abide by the base 10 rule | Bikerman wrote: |
The question was perfectly serious - not at all tongue in cheek. If you use different number bases within each expression, for example:
10 +10 +10 = 6 is easy (10 base 2, and 6 base 7 or greater).....
However, I'll abide by the base 10 rule |
10 + 10 + 10 = 110 in base 2. 6 is meaningless. | jeremyp wrote: |
| Bikerman wrote: | The question was perfectly serious - not at all tongue in cheek. If you use different number bases within each expression, for example:
10 +10 +10 = 6 is easy (10 base 2, and 6 base 7 or greater).....
However, I'll abide by the base 10 rule |
10 + 10 + 10 = 110 in base 2. 6 is meaningless. |
Come on Jeremy - read what I wrote ; | Quote: |
| If you use different number bases within each expression |
| Bikerman wrote: |
| jeremyp wrote: | | Bikerman wrote: | The question was perfectly serious - not at all tongue in cheek. If you use different number bases within each expression, for example:
10 +10 +10 = 6 is easy (10 base 2, and 6 base 7 or greater).....
However, I'll abide by the base 10 rule |
10 + 10 + 10 = 110 in base 2. 6 is meaningless. |
Come on Jeremy - read what I wrote ; | Quote: | | If you use different number bases within each expression |
|
But you have to find a way of expressing the base of each number if you're going to do that (without using digits). | jeremyp wrote: |
| But you have to find a way of expressing the base of each number if you're going to do that (without using digits). |
Touchι...OK, you got me Hi jeremyp
| jeremyp wrote: |
| (0! + 0! + 0!)! = 6 |
Yeah, that's the crucial step. Well done, most folks don't think of using the factorial operator! 9 / (sqrt)9 + (sqrt)9 = 6?
(6 * 6 / 6 = 6) or (6 / 6 * 6 = 6)? 2 * 2 + 2 = 6?
3 * 3 - 3 = 6?
(sqrt)4 + (sqrt)4 + (sqrt)4 = 6?
7 - 7 / 7 = 6?
I was just wondering, would there be multiple answers for each part? Because in certain cases, it is possible to end up in the same result using different operators.
5 / 5 + 5 = 6?
I don't quite understand the answer for 0. Can't think of an answer for 0, 1, 8 and 10 yet... 
The answer for 0 uses an operator called the 'factorial' (commonly written as !).
Factorial 3 (3!), for example, is 3*2*1. Factorial 5 (5!) is 5*4*3*2*1
Normally, then, we can say that n! = 1*2*3*...*n
Now, the trick is in working out what factorial 0 is.
For an explanation see; http://mathforum.org/dr.math/faq/faq.0factorial.html Wow! That really opens up my mind... Thanks for the information! =)
Um..
8/(cube root)8 + (cube root)8 = 6?
Hello Folks
So far you've seen straight forward solutions for all the numbers but 0, 1, 8 & 10.
There is also a straight forward solution in the case of 8 (i.e, doesn't require the factorial(!)) operator.
However, as far as I know, the solutions for 0, 1 & 10 do require the use of the factorial (!) operator. At least, I'm pretty certain that factorial is needed for the number 0, as it's the only simple operator that will make something out of nothing (i.e. turn 0 into a non-zero number). May be someone out there can do better!
BTW, thanks Bikerman, for your explanation and reference for the factorial operator - I had totally forgotten about the gamma function! Let me just add another explanation for 0! = 1: Since n! is the product of the first n integers, 0! should be the product of the first 0 integers, i.e. the product of no numbers at all. If I were to ask what is the sum of no numbers, everyone would agree that this is zero, the neutral element for addition (0+n=n); indeed n x 0 = 0. Similarly the product of no numbers is 1, the neutral element for multiplication (1xn=n): in fact n^0=1. So 0!=1. Of course, the need for 0! x 1 = 1! = 1 confirms this.
| mk12327 wrote: |
2 * 2 + 2 = 6?
3 * 3 - 3 = 6?
(sqrt)4 + (sqrt)4 + (sqrt)4 = 6?
7 - 7 / 7 = 6?
I was just wondering, would there be multiple answers for each part? Because in certain cases, it is possible to end up in the same result using different operators. |
I should have mentioned that there of often multiple solutions to these problems.
Have fun! Using factorial in the solution, answer for 1 seemed so much easier... LOL...
(1 + 1 + 1)! = 6
Ok, this looks abit overkill but here goes my solution for 10:
((sqrt)(10 - 10 / 10))! = 6
Hi mk12327
| mk12327 wrote: |
Ok, this looks abit overkill but here goes my solution for 10:
((sqrt)(10 - 10 / 10))! = 6 |
It does look like overkill, but it is my own solution, i.e. I haven't yet found any simpler solution. May be someone else can...
Hey, this looks a lot neater in LaTeX:
Too bad we don't have LaTeX available on Frihost forums (yet!) I have raised the matter and Bondings has agreed to look at it....can't say fairer than that.
What do you call a mathematical operator? For example, if we consider trig functions and the floor function (pretty-looking in tex) as mathematical operators, then for the problem with the 8's an answer would be:
| Code: |
| 8 - floor( csc 8 + csc 8 ) |
| nilsmo wrote: |
What do you call a mathematical operator? For example, if we consider trig functions and the floor function (pretty-looking in tex) as mathematical operators, then for the problem with the 8's an answer would be:
| Code: | | 8 - floor( csc 8 + csc 8 ) |
|
The floor function is often known as the greatest integer function. It's represented like [x] except that there's a line down the bracket. To go the other way (rounding up), you'd simply put -[-x]. | nilsmo wrote: |
What do you call a mathematical operator? For example, if we consider trig functions and the floor function (pretty-looking in tex) as mathematical operators, then for the problem with the 8's an answer would be:
| Code: | | 8 - floor( csc 8 + csc 8 ) |
|
Cool solution! That's the first really inovative approach I've seen in a while. Too bad you lose points by not presenting the answer neatly in LaTeX.
OK, first of all I should have said I didn't invent this problem. A friend of mine told me about it, saying it had originally been set as a test in a Japanese secondary school(!)
Since it's up to me to set the rules here, I would say that it's in the spirit of the problem is to avoid using numbers and letters, i.e. the only functions we are allowed are those represented by symbols (i.e. from 
or 
). But I think your use of functions is so neat that they should be allowed in a "grown up" (X-rated?) variant of the problem.I like the related 4 4s problem, where you use 4 4s to come up with every number 1 through 100. For example, 4^4 - 4/4 = 63.
Good times.
Hi PatTheGreat42
Thanks for this new problem!
How about this for starters:
4/4 + 4 - 4 = 1
(4 x 4)/(4 + 4) = 2
(4 + 4 + 4)/4 = 3
√4 + √4 + 4 - 4 = 4
√4 + √4 + 4 /4 = 5
4 + 4 - 4 + √4 = 6
4 + 4 4/4 = 7
4 + 4 + 4 - 4 = 8
4 + 4 + 4/4 = 9
4 + 4 + 4/√4 = 10
Can one really reach 100?
Good luck everyone!
For the 4 4s problem, why not we make it new thread for it? Sounds interesting to me... haha...
Hi Bikerman
| Bikerman wrote: |
| I have raised the matter and Bondings has agreed to look at it....can't say fairer than that. |
Is there any news on the LateX issue?
Regards | infinisa wrote: |
Hi Bikerman
| Bikerman wrote: | | I have raised the matter and Bondings has agreed to look at it....can't say fairer than that. |
Is there any news on the LateX issue?
Regards |
I'll raise it again | infinisa wrote: |
It does look like overkill, but it is my own solution, i.e. I haven't yet found any simpler solution. May be someone else can...
Hey, this looks a lot neater in LaTeX:
|
I thought of a much more simple solution. Generally when a base is left off of a logarithm, the base is assumed to be 10. Therefore:
(log(10)+log(10)+log(10))!=6.
Although I do not know if that is valid for this problem because if I were to write this in a real life problem, I would most likely write: log_10(10).... even though it's generally assumed that the base is 10 (or "e" if written as ln()) if it's not written...
