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I thought that dog was kind of cute.
Is that dog in the vid yours, Bobby, and is that you speaking? No one other than the dog's owner would ever call it cute!
The product of the two inner digits (7*3) less the sum of the two outer digits (5+3) = 21-8 = 13.
Hi ZHero,
L
You'll have to fill me in, ZHero, 'cos I don't see a trend (other than increasing powers down the lists). >blind<
Doesn't it look like "The Cheese" to you phro
?
It would have been holey Swiss cheese - if the ugly mutt in that vid had sunk its teeth into it - instead of the real cheese that Tulip had.
5681 = (13) × (13 + 3 × 1 + 1 × 3) × (13 + 3 × 3 + 1 × 1)
Can You Notice Any Trend in The Above Numbers?
A trend of one exists between the size of the group of numbers above the first dotted line and that below it: the first group contains 50% more lines than the second group and has a "+" as each line's last operator instead of a "-".
I just noticed a faint, indiscernible orange splotchy blotch after each green-coloured number. Wonder if they mean anything.
All unique sequential numbers in your magic square, ZHero!
I constructed a magic square by hand like you did, Bobby, and drew up a 5577 x 5577 unitary magic square (ie, all digits are "1", and are non-unique and non-sequential). The verification process (ie, to check the row, column and diagonal sums) took quite a while because I chose to sum all those numbers in my head, but I finally got there! And it checks out!
I'm rather loath to give out my email address, so I'll pass on getting you to send me that file, Bobby. Sorry, 'bout that, after you went to all that trouble!
No...I didn't think it was a built-in function you were using. That is why I thought you'd been having fun trying to get the thing to work.
If it's mathematica-intensive, it's goodnight me. If it's ugly and something to be ashamed of, then it's goodnight it.
I suspected Mathematica was at the bottom of this...you must have had a bit of fun setting it up to the stage it's at!
5499
(5!/4)/√9+√9 = 13
No good me trying to learn how, then.
You just need fast hands.
You mean fast fingers on the ends of your hands?
Completely done by hand. Everything sums to 5408.
How on earth can you do something like that by hand? Or by anything else, for that matter??
Ok. Had to look up induction now (Wikipedia, as it's not on MIF). You keep on sending me to the dictionary! Just a quick look only, but I sort of get a rough idea about it. Way beyond me at my unlearned stage.
Hi Bobby,
Thanks for your explanation. Looks like the answer probably is .
Hi Bobby,
Had to look up MIF's definition of coefficient, as I'd either forgotten it or wasn't taught it. The former is the more likely!
I think the answer is .
Glad to know ya!
The pleasure is mine. I enjoy your input and helpfulness.
I was always far too sports-mad (tennis, table tennis, Aussie Rules football, squash) to be interested in learning to play a musical instrument - which disappointed my mother greatly (she played the guitar and sang beautifully).
For a while I got a bit of an interest going in playing the synth, but after I'd tried out all the buttons that interest evaporated.
I like unaccompanied voice harmonies and sing in an a cappella male quartet. That takes up a lot of my time as I lead the group.
I prepare sheet music for the others on a midi-compatible software program, and burn a CD for each of them with their part in the right channel and the other 3 parts in the left...to help with learning their part. For the sounds I use a midi tone generator that the sheet music program can talk to.
We're just an amateur group and sing at our church camps and conventions, and sometimes at weddings. Good fun!
Thanks also for the music.
That was something I did to help one of my relatives with their school project on synthesizers, as an example for how synths can be used to bring a variety of sounds into a performance.
The quantizing feature was very helpful!
Hi Bobby,
I guess I could just substitute for n to get one formula, which avoids having to do any plugging in.
Also your challenge has been solved over at club 13. Will you check the solution and verify it's uniqueness.
Done!
Hi phrontister!
sorry i just couldn't resist myself from posting 5239 again!
No probs...it is good for my mind to be broadened.
But you didn't post a new number!
Some more primes with that number 5239: 2, 3, 5, 23, 29, 53, 523.
Also,
52 = 4 x 13
52 - 39 = 13
52 + 39 = 7 x 13
5 + 2 - 3 + 9 = 13
5 + 2 + 3 + √9 = 13
Hi Bobby,
Correct! The solution is unique.
5265 (catchup):
-52 + 65 = 13
5 + INV2 * 6 + 5 = 13
5278
-52 + 78 = 26 = 13 + 13
52 + 78 = 130
The product of the outer two digits - the inner two digits (27) = 13
What happened to you at metamorphosis?
Work's been very hectic these past few weeks and I just can't fit everything in. Something out of maths problems and word games had to give, and the extra variety in maths has been the drawcard.
I enjoy word games, and metamorphosis is a good challenge. Must get back into it again...when I can.
Thanks for letting me know...I'd forgotten about it.
Hi Bobby,
Thanks for your explanation...I even made sense of most of it!
I had to ignore "methods of interpolation" and "I curve fit that table", because they went over my head.
We had the same major goals, but went about it a little differently.
Here's a better explanation of my method.
5239
FIVE - TWO + THREE - NINE = 5239
Each letter in the puzzle stands for a different digit from 0 to 9. Find the letter values.
I'll try to think up some questions before I see the production.
Publish your performance on YouTube. Some greats started that way.
Maybe not with dancing, though...although you could break new ground. That is, if you don't fall over!