Sarah's Key

[Ps 27:1]

Steve




I have read and waited to the end for the solution and I am as confused with your solution as I was at the beginning with the problem. Sorry if I am thick in understanding the solution to your problem, that is assuming a sloution exists. There does not have to be a connection with the 9 unary system which is a subset of the decimal system conforming to the number of our fingers (digits). As you know our tenth digit is a single object but in the decimal system is representd by two symbols; 1 and 0 where 0 is a symbol and not a digit. I expect there is a convention for for numbering the fingers on the hand replacing left hand with one set of digit numbers and right hand with another set of digit numbers and replacing the index finger or thumb identification etc. for a number.

Your solution to me looks like fitting a square peg into a round hole, except this time, you fitted a round peg/shaft into a square hole. I cannot help myself recalling an expression which the more crude would use to say what has been done.

I look forward to more enlightenment.


David

Hi David,

Zero (0) is a symbol(numeral) and a digit. http://en.wikipedia.org/wiki/Numerical_digit and http://en.wikipedia.org/wiki/Zero

A round shaft in a cube would be a good description of the solution. If you knock the cube over, you will see the hole (0) in the middle.

Remember, this is a riddle (fun ? puzzle), not a mathematical theorem (provable fact).

Steve

[Ps 27:1]

Yeah, I can see that. His "fox in the box" hit the bullseye.

Was the pun intended?

Well, I still don't see what's so "nice" about that "home" for the zero. What does it really add to our understanding?

Seriously? I get that all the time concerning gematria, so it seems a bit strange coming from you. I thought we were just having fun, man. It's not like we're solving the universe here.

The form of 505 matches you solution, but only in hindsight. My solution of a 5 x 5 square of zeros with the 3 x 3 magic square centered to give a wall of 16 zeros surrounding fits with that number too.

Well, you know what "they" say about hindsight? (20,20) Wouldn't your solution go with 050? If you believe it, then it will be.

But "fox in a box" was pretty good!

You do know that fox in English gematria = 6+60+600= 666? Have you ever seen a silver fox in the wild? I have once and the circumstances surrounding that day and night became indelible in my mind. But that is for another thread.

As for an "algrithm" to list out the number 3142 from the square, the simplest is probably "list out the numbers 3142 from the square." That's probabily the smallest string that would satisfy you demands, and if so it would technically be the answer to your riddle. It will be interesting to see if your algorythm can be expressed in a shorter string.

Have a great day.

Richard

One of my original stipulations was that you had to use all 9 digits. My original instructions were also vague. So let me try again. Come up with 3142 using simple arithmetic and all of the magic square. You can +, -, x, /. No exponents, etc. The simpler, the better. If no one wants to try this, I will post my solution promptly.

Thanks and the same to you,
Steve

[David M]

Solution to obtaining 3142 from using all the digits 1 2 3 4 5 6 7 8 9 is as follows:


Maybe I struck on this by the luck of choosing certain numbers early on.

I knew that 9 x 8 x7 x 6 = 3024 and adding this to 5 x 4 x 3 x 2 = 120 gets you to 3144 and taking one off got to 3143 which is close but not good enough.



I tried a few more combinations to come up with 9 x 6 x 5 x 4 x 3 = 3240 which gets you within 98 of the required target.

Can you get 98 from 8, 7, 2, 1 that remain?


The solution to this came as ((8 -1 = 7) x 7 = 49) x 2 = 98 so I guess I had done it.


The final solution is therefore:

(9 x 6 x5 x 4 x 3) - ((8-1) x 7) x 2 = 3142



David

[Richard Amiel McGough]

Solution to obtaining 3142 from using all the digits 1 2 3 4 5 6 7 8 9 is as follows:


Maybe I struck on this by the luck of choosing certain numbers early on.

I knew that 9 x 8 x7 x 6 = 3024 and adding this to 5 x 4 x 3 x 2 = 120 gets you to 3144 and taking one off got to 3143 which is close but not good enough.



I tried a few more combinations to come up with 9 x 6 x 5 x 4 x 3 = 3240 which gets you within 98 of the required target.

Can you get 98 from 8, 7, 2, 1 that remain?


The solution to this came as ((8 -1 = 7) x 7 = 49) x 2 = 98 so I guess I had done it.


The final solution is therefore:

(9 x 6 x5 x 4 x 3) - ((8-1) x 7) x 2 = 3142



David

Good work David! I was going to try something along the same lines, but I don't think it can be correct because Steve said that it has to be "an algorithm, simple enough for a child to understand." Here's the conditions:

Okay, second part to this riddle. PI is one of the most important numbers in mathematics, agreed? Give an algorithm, simple enough for a child to understand, that uses all of the magic square to get 3142. (PI to the nearest thousandth ignoring the decimal.) Kind of like a pin number.

Bonus points if you can extend the string, 314159...

Steve

I think you gave a formula, not an algorithm.

[David M]

Richard

I had forgotten about that first challenge and was concentrating on the last post which said;

One of my original stipulations was that you had to use all 9 digits. My original instructions were also vague. So let me try again. Come up with 3142 using simple arithmetic and all of the magic square. You can +, -, x, /. No exponents, etc. The simpler, the better. If no one wants to try this, I will post my solution promptly.

Maybe the first challenge should continue. I am not sure where to start with algorithms for calculating pi. It is the ratio of the circumference to the diameter of a circle and therefore will involve higher mathematics that I have long forgotten how to use.

David

[Ps 27:1]

The solution to this came as ((8 -1 = 7) x 7 = 49) x 2 = 98 so I guess I had done it.


The final solution is therefore:

(9 x 6 x5 x 4 x 3) - ((8-1) x 7) x 2 = 3142

Hi David,

That is the approach most of us would probably take, myself included. That is why I said ,"simple enough for a child to understand." I had students in high school that had a hard time with order of operations. I was just playing around when I "accidentally" discovered my solution.

Blessings,
Steve

[Ps 27:1]

Good work David! I was going to try something along the same lines, but I don't think it can be correct because Steve said that it has to be "an algorithm, simple enough for a child to understand." Here's the conditions:


I think you gave a formula, not an algorithm.

Now I'll have to psychoanalyze you so I can guess what kind of answers you are expecting.

I think you're starting to have me all figured out.

Steve

[Ps 27:1]

Richard

I had forgotten about that first challenge and was concentrating on the last post which said;




Maybe the first challenge should continue. I am not sure where to start with algorithms for calculating pi. It is the ratio of the circumference to the diameter of a circle and therefore will involve higher mathematics that I have long forgotten how to use.

David

Whoaa! Stop right there. Forget all that for now. Just think of a simple way to get to 3142 using all of the square. I'll give some hints if you're stumped. I don't want to send you on a wild goose chase and then have you mad at me.

Peace brother,
Steve

[Ps 27:1]

Not that anybody cares, here is my solution:

Those who have played with a compass, probably know that the radius of a circle can be used to divide the circle into 6 congruent arcs. Thus you can construct 6 circles around one. Also, 6 congruent coins will perfectly surround a seventh.



I would sometimes ask my students why we have 7 days in a week. Of course, the most common reply is the creation story. I would say that was a possibility (I had to remain objective, because it was a geometry class in a public school ) and then I would tell them that the 6 around 1 is a prominent feature in nature and then we would construct the above image.

When I was playing around with the 3x3 magic square, I wondered what would happen if I summed the "7 days". This is what I did:

159
258
357
456
555
654
753
-----
3192

I saw that the sum was exactly 50 more than 3142 and that 555 was really inconsistent with the other strings. All the other strings are made by lines passing through the center, but 555 was not. It seemed serendipitous that the previous riddle provided the solution for the 505 string to now be linear and making the sum fit PI.

As you can see, it should not be hard to give a kid instructions on how to come up with the list of strings. I'll leave that up to you.

Cheers,
Steve