2,7,8 puzzle and 6 N puzzle

So here is a puzzle I herd a while back and I started thinking about it in class earlier today. The puzzle is "using the numbers 2, 7, and (0-9) make an equation that equals 8. You may use +,-,*,/,decimals, (), and powers." For example, for equation 3 I could have 7+3-2=8. Post what equations you get for 0 through 9. Extra credit for 11 through 29, additionally you may use any mathematical function you know of (I know there is a solution for 0-9, but I don't know/haven't figured out some for everything from 11-29).


So here is a second puzzle I just herd from my friends, I am not sure it is fully do-able, and I couldn't find the puzzle on line to confirm. So using exactly 6 N's per equation (where N is any integer besides 0) you must make an equation which equals 0-9. For example the solution to one could be N*N*N/(N*N*N)=1. You may use any mathematical functions you wish, however any that involve a number must have an N as the number and that N counts toward your 6. (for example Log would need to be Log base N and would leave you 5 more N's. Also integrals and derivatives require a dN and this N counts toward your 6). My friends and I could only get 0-7 so we think there may be a rule we are missing but I figured I would post this too and hopefully find out what I am missing.

Good luck

Simple starts:

((N/N)+(N/N)) x (N/N) = 2
(N/N)+(N/N)+(N/N) = 3

Puzzle 1

(2⁰ + 7) = 8
(2-1) + 7 = 8
7 + 2/2 = 8
3 you already gave
2^7-4 = 8
...

Puzzle 2
(N-N)+(N-N)+(N-N) = 0
...

Dammit, 4tran, stop grouping the Ns together!  That's limiting our thinking.  What the fuck is the (or an) N solution for 4?

>>4
?  I'm not doing anything you're not already doing.

(N+N)/N+(N+N)/N = 4
(N+N+N+N+N)/N = 5

Here's a quasi-proof demonstrating the limits of only using 6 Ns (if you cannot use any other numbers):

To get a number which does not contain 'N' in it at the end, you must divide by N at least once (to cancel out the Ns in the numerator).

This takes up at least one N. For any multiplication in the numerator, you must have another N in the denominator. Thus, any time you multiply by N you need an N in the numerator and one in the denominator. Thus multiplication uses 2 Ns and does not increase the coefficient, so it literally "wastes" an N when we're trying to find the maximum value.

e.g.: N/N = 1. N*N/N = N so we need an extra N in the denominator: N*N/(N*N) = 1. The coefficient/solution has not been increased.

For purposes of finding the maximum value we can achieve, multiplication is non-productive.

Subtraction and division make things smaller, so clearly they are not productive.

For an even worse reason than multiplication, powers (and in the middle, parenthesis do the same) are problematic and useless (N^N would require N Ns in the denominator to cancel out and get us a non-N-based solution).

An integral is pointless because, to get a numerical value, we need a range for the integral to operate over, but we can only use Ns. We have two possible indefinite integrals:

int(dN) and int(NdN).

int(dN,N=a..b)=b-a. But b and a must be something using Ns for our purposes. And, as stated above, subtraction is worthless for our goals. We need a maximum b and a very tiny a. a cannot be smaller than 0 (N-N) and b cannot be bigger than 6 (N+N+N+N+N+N) for discussion ruling out *,/,- earlier. And, to get the 0, we need the large number to use no more than 4 Ns, so we have 4-0=4. But we've used an N in the integrand itself, so we only have 3 Ns to use excluding integrand and the (N-N) to get 0. So we actually have int(dN,N-N,N+N+N)=3.

Through similar work, we can see that int(NdN) can give us nothing good at all: we have N^2/2 from a to b, and clearly we need to minimize a, which means either N-N for 0 or N/N for 1. We're then left with only 2 Ns to use for b. The max to use there (and remember it cannot have any Ns in it or our solution will have an N in it) is N/N. So the max attainable through int(NdN) is int(NdN,N-N,N/N)=1/2. This is horrible and not even a whole number. I think the largest whole number attainable is 0: int(NdN,N-N,N-N).

Any derivative using Ns must culminate in a whole number, which means the expression we're deriving must be linear. So we need to use only linear operators to attain the maximum. Our derivative operator is d[f(N)]/dN. So we have 5 Ns to use for f(N). The max here is N+N+N+N+N. d[N+N+N+N+N]/dN = 5. 5 is the max. Other combinations are less productive, such as [d[N*N+N*N]/dN]/N = 4.

I'm growing tired of this, let's just get to addition. (N+N+N+N+N)/N is clearly the largest number. This is 5.

I'm not going to exhaust every possible mathematical function, but I can assure you that the most productive (and simple) way of attaining a large number is addition (with the required single division on bottom): (N+N+N+N+N)/N=5.

I'd looooooooove to see how you got 6 and 7, faggot.

>>7
6=((N+N+N)/N)! * N/N
7=((N+N+N)/N)! + N/N

As I said, I am only stuck on 8 and 9 however at this point I am assuming they cannot be done unless I am given another special function (free derivatives or 10^X or something)

>>8
Oh, goddamn factorials. I forgot about those. But factorial grows faster than any other unary function I know of, so that should really be the upper bound. Unfortunately, the upper bound it would give is 120 [(N+N+N+N+N)/N]!

Thus, my "proof" was worthless. :(

Has anyone tried using inverse trig functions?

arccos(-N/N) / arctan(N/N) + N/N = 5

I didn't have any luck producing 8 or 9.  Maybe someone else might be able to use them for something.

Any mathematical function? Here guys, you can use mine if you want.

f(N) := m, where m is any number you damn well please

/thread

>>9
keep in mind that we can use the factorial multiple times.  i.e. (((N+N+N+N+N)/N)!)!=120!=a very large integer.  There is no upper bound.  Of course, factorial might not be allowed, since the problem doesn't specify POSITIVE integers.

>>12
the factorials work just fine with a negative N.  Keep in mind that the equation must work for all non 0 ints (the non 0 is so that you can divide by N) ((N+N+N)/N)! will be 3! regardless of what non 0 N value you have as the N's will cancel out.

>>11
Yeah this is kind of silly.

Even if you limit it to established functions, it's not that hard, e.g.:

8 = F(((N+N+N)/N)!) * N/N,
9 = F(((N+N+N)/N)!) + N/N,
where F(n) is the nth Fibonacci number.

>additionally you may use any mathematical function you know of

Haha! I know this one good function. It's called constant function .

Let F(x) = 8

Problem solved for every possible combination of numbers!

>>9
The fastest growing unary function I know is the Ackermann function (defined as the binary Ackermann function with the same input for both variables).

>>13
I last recall factorials being generalized to gamma functions to include all real numbers.  This would result in the limits of the factorials of negative integers to be +- infinity.

>>16
Try reading >>13 again.

>>17
What?  I did read >>13, and that's why I was commenting on it.

I claim that the factorial function is singular for negative integers.

N! = integral[t^N e^-t, t, 0, infinity]

Thus, [(N - N - N)/N]! = (-1)! phails.

As >>13 pointed out, whether N is positive or not doesn't matter, but it does matter if the input to a factorial is positive or not.

For 8 you can use ((N+N)/N+(N+N)/N)!!=8  (note, !! is the double factorial, not the factorial of the factorial)

If we can use set theory, then 9=({N,{N},{{N}}})*({N,{N},{{N}}}).  Of course, since we can write 0={}; 1={{}}; 2={{},{{}}}; 3={{},{{}},{{{}}}}; ad infinitum, it would render the exercise rather trivial.