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Posts posted by bonanova


16 3 2 13
5 10 11 8
9 6 7 12
4 15 14 1
The ancients associated magic squares with the astrological planets.
the 4x4 version was associated with Jupiter.
But you probably had something else in mind?

All of them. There is only one, which is 2. All other even numbers are divisible by 2 and therefore not prime. The question doesn't specify that the solution must be evenly divisible by 5, therefore ALL numbers are divisible by five, including 2.
Hmmmm.. well, ok.
But I've edited the question now, hopefully asking what I had intended to ask.
See if your answer changes ...

I think this answer
doesn't work: when you get to the South pole, how do you run West?Yep. I presume you mean 1 mile north of south pole?But this answer:
does: for example, any point on the circle (1 + 1/2pi) miles from the South Pole.There's an infinite number of circles around the South Pole where he could have started.After going South 1 mile, you're (1/2pi) miles from the Pole,
which allows you to run West 1 mile [1 lap of a 1mile circumference circle]
and be able to go a mile North to the starting point.
As Martini noted, there is an infinite number of starting distances:
1 + 1/2Npi miles North of the South pole where N is any positive integer.
N is then the number of circular laps in your westerly mile.
e.g. N=5280  you'd run 5280 laps around a 1foot circumference circle.
Here's a counter question  why can't N be negative?
i.e. start closer than a mile  you could still do N laps
 1
 1

Choose the best answer and say why.
Which of the even prime numbers [excluding 2] are [evenly] divisible by 5?
[1] all
[2] some
[3] none
[4] the question has no defensible answer.
p.s. Martini: I assert this is a logical not mathematical question.
Edited for "clarity". [meaning that I originally screwed up what I wanted to ask .. ]
I do not choose [4].

There are digits as high as 9, so the basis would have to be 10 or greater.You got the hint (bases)But you can't go from twentysomething to fortysomething by adding 7 with a basis that high.
I'm curious to see how this turns out.

If I understand the "parity" constraint, the idea is to keep neighbors from being nearest or nextnearest in value.
Neighboring numbers must differ by 3 or more in value.
I broke the problem down this way.
First note that 1 and 14 have 11 allowable neighboring numbers; 2 and 13 have 10; the others, 3 thru 12, have 9.
The 2 most central squares of the grid have 8 neighbors, and their outside neighbors have 6 neighbors.
The others have fewer: 3, 4 or 5.
So I start by placing 114 in the middle, and 132 as their outside neighbors.
The numbers with the most flexibility are placed in the most demanding squares.
So we start with this for the middle row:
x  13  1  14  2  x
Then I try to keep symmetry by placing the remaining numbers in pairs:
312, 411, 510, 69 and 78. Start with 312 in the outside squares:
3  13  1  14  2  12
and keep going, making sure the lines connecting the added pair pass thru the middle of the grid:
11
3  13  1  14  2  12
4
1011
3  13  1  14  2  12
45
10611
3  13  1  14  2  12
495
106118
3  13  1  14  2  12
7495

that's why i put the smiley there ... a little humor.

yeh mon, lay it on us... [it's not octal at least ... but what?]

How about ...
5 4 3 2 8  can't show the tile boundaries easily, but it should be obvious.
4 6 7 1 9
3 7 0 4 2
2 1 4 1 6
8 9 2 6 7
Here's how:
If a number appears an odd number of times, it has to be on the diagonal.
Five numbers do that: 0[1], 1[3], 5[1], 6[3], 7[3]
Upper left UL corner has to be top or left number on a tile.
Five possibilities: 1[on 3 tiles] 5[on 1 tile] 7[on 1 tile]
Lower right LR corner has to be bottom or right number on a tile.
Three possibilities: 6[on 1 tile] 7[on 2 tiles]
Start eliminating the possibilities. In the order mentioned above,
Start with UL corner:
[1] 16 tile.
Requires a 6 beneath the 1.
No 6 appears on the top or left of a tile. Impossible.
[2] 19 tile. Requires a 9 to right of the 1.
No 9 appears on top or left of a tile. Impossible.
[3] 14 tile.
Requires a 4 to right of the 1. Place the 467 tile there.
That puts a 6 on the diagonal. OK.
Also requires a 7 right of the 6. Place the 704 tile there.
That puts a 0 on the diagonal. OK.
Bottom right number must now be 5 or 7. No 5 on bottom or right of a tile, so place 267 tile there.
Requires a 5 above the 6.
No 5 appears on bottom or left of a tile. Impossible.
[4] 543 tile.
Requires a 4 right of the 5. Place the 467 tile there.
That creates the diagonal6. Now the only possible BR number is 7. Place the 267 tile there.
That requires a 6 above the 7. Place the 16 tile there.
A 2 must be above the 6. Place the 29 tile there.
A 9 must be left of the bottomrow 2. Place the 19 tile there.
The remaining diagonal number is 0. Place the 704 tile there.
4 must be right of the 0. Place the 14 tile there.
Finally place the 28 tile and the 328 tile in their slots.
Check for symmetry.
Done.
[5] 704tile.
Requires a 0 right of the 7.
No 0 appears on top or left of a tile. Impossible.
Start with LR Corner
[6] 16 tile.
Requires a 1 above the 6.
No 1 appears on bottom or right of a tile. Impossible.
[7] 467 tile.
Requires a 6 to left of 7. Place the 16 tile there.
That places a 1 where a 4 has to be. Impossible.
[8] 267 tile.
Hmm... looks exactly like [4] above. It is. Done.

Antispoiler: you mean: the grid is symmetric about its [upper left  lower right] diagonal?
If so, no need to peek at the spoiler above...

brooms with corners? strange looking broom; but if you say ..."Well, there could b rooms w/ corniers.btw there's a typo in 'broom' .. [extra space]

maybe ?
2 for each girl [4x7=14],
4 for each big cat [4x7x7x7=1372] and
4 for each small cat [4x7x7x7x7=9604]
10990 in all

It was quiet at Morty's last night ...
Until Matt the Mathematician proposed a game.
Alex and Davey would both put their billfolds on the table,
and Matt would count the money in each.
Whoever had more would forfeit his money to the other player.
"Think it over," Matt said, and he ordered another frosty one.
Alex thought to himself,
"There's no reason to believe I have more money than Davey
has, or less, for that matter; so my chances of winning are 50%.
Now if I lose, I'll lose whatever's in my wallet.
But if I win, it'll be because Davey has more money than I.
So on a tossup bet, my winnings are more than my losses."
"I'm in," he finally announced.
And Davey thought the same.
He figured his chances of winning were as good as Alex's,
and he'd either lose the amount in his wallet or win
an amount greater than that.
"You're on," said Davey with a smile.
If the boys were thinking clearly, how could the game favor both of them?
Or if not, where is the flaw in their reasoning?

proving that ...
[1] the smallest number not specifiable in fewer than 23 syllables can be specified in 22 syllables.
[2] there are only a finite number of numbers.

Bravo, Writersblock.1,177,777
one million, one hundred seventy seven thousand, seven hundred seventy seven = 23 syllables
Onemillion onehundred seventy seven thousand seven hundred seventy seven = I count 22.
But I realize I am wrong with 7,777,771. It should be 1,777,777.
Onemillion seven hundred seventy seven thousand seven hundred seventy seven = 23.
1,777,777 is the smallest number not specifiable using fewer than twentythree syllables.
At least, no one has come up with a smaller number. So let's say it is.
You get the prize.
O wait. This is supposed to be a paradox.
ummm, just for the heck of it, count the syllables in red, above.
If the red words specified your answer, then ....

ROTFLWhy do I feel like the bastard child of Karl Childers and Forest Gump?
I'm with ya ... I don't know who Karl Childers is...

For any nonzero, finite number, say 13, for sake of discussion, then usually:
[1] 0/13 = 0
[2] 13/0 = infinity
[3] 0/0 = undefined [that is, could be 242, for example]
[4] infinity/infinity = undefined [could be 0.327, for example]
[5] 0 x infinity = undefined [could be 69, for example]
But it turns out that both zero and infinity have different flavors, called orders.
There are 1storder zeros, and 1storder infinities, as well as 2nd, 3rdorder, etc. of both.
In these cases, the higher order predominates:
[3a] 0[2]/0[1] = 0; 0[1]/0[2] = infinity
[4a] infinity[2]/infinity[1] = infinity; infinity[1]/infinity[2] = 0.
[5a] 0[2] x infinity[1] = 0; 0[1] x infinity[2] = infinity.
where 0[n] means nth order 0 and infinity[p] means pth order infinity.
You can see what that means by changing one of these zeros a wee bit, then inching your way back.
Let x be a small number  say 1/100.
[6] x/x=1. no problem, since x isn't zero. Let x inch its way to zero, the ratio stays at 1.
[7] 2x/x = 2. no problem, since x isn't zero [yet]. Let x > 0, the ratio stays 2.
That's why 0/0 is usually thought of as undefined: you could have used any number instead of 2.
But look at
[8] x^2/x = x. now the ratio is x. Let x >0, and the ratio [always = x] >0 as well.
What happened? both numerator and denominator > 0, so the ratio goes to 0/0.
How come it's not undefined? How come in this case we know it's zero?
The reason is x^2 becomes a 2ndorder zero: 0^2 so to speak; it's stronger than just 0.
Same thing about the infinity you get by dividing by x^2 instead of by x as x>0.
There you get a stronger infinity  a 2ndorder infinity  infinity^2 so to speak.
There is a rule called L'Hopital's rule that gives you these answers for [6]  [8].
Read more about that here: http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule
Basically you step away from the dividebyzero situation and evaluate the division using a legal small number, then see what happens when it approaches zero, and that gives you a defined answer, for that particular case.
L'Hopital's rule can make some paradoxical problems intuitive.
Absent some expression like [2x/x] >2, or [x^2/x] >0, or [x/x^2] > infinity, you're just left with undefined and you don't try to prove too much with expressions that contain infinities in the numerator or zeros in the denominator.

Interesting.If they are bald they have no hair to be judged by. They can not have the same amount of hair on their head0 does not equal 0.
But let's keep this a secret, OK?
Word gets out, imagine the recall of all those text books.
Would be devastating.

The inscriptions on Gold and Silver agree, so they must have the same truth value.
By conditions of the problem, Lead has the opposite truth value.
Gold and Silver False means
[1] Ring is in Silver
[2] Lead is true > Ring is in Lead.
Contradiction
Gold and Silver True means
[1] Ring is in Gold or Lead
[2] Lead is false > Ring is not in Lead.
Ring is in gold box.

That's it.
Here's my method: [care to share yours?]
The number has to end in 9.
Looked brute force for small numbers.
59 and 119 were promising, but no cigar.
Then looked for agreement among
39 + multiples of 40,
69 + multiples of 70 and
89 + multiples of 90
Smallest one was 2519.
Still think of this as kind of brute force.
Maybe there is no elegant solution.

That's it.

Considering only positive integers, 7 is unambiguously specified by the phrase the smallest number not specifiable using fewer than two syllables.
What is the smallest number not specifiable using fewer than twentythree syllables?
How many syllables does the smallest number not specifiable using fewer than twentythree syllables have?

I just found a number with an interesting property:
When I divide it by 2, the remainder is 1.
When I divide it by 3, the remainder is 2.
When I divide it by 4, the remainder is 3.
When I divide it by 5, the remainder is 4.
When I divide it by 6, the remainder is 5.
When I divide it by 7, the remainder is 6.
When I divide it by 8, the remainder is 7.
When I divide it by 9, the remainder is 8.
When I divide it by 10, the remainder is 9.
It's not a small number, but it's not really big, either.
When I looked for a smaller number with this property I couldn't find one.
Can you find it?

A prime, N, is a number with no factors other than [1,N].
Can the arithmetic mean of two consecutive prime numbers be itself a prime number?
75 Bands
in New Logic/Math Puzzles
Posted
How about ...
21. Doors
22. Cars
23. Led Zep
24. Kiss [the um, interesting couple under the traffic light]
25. Madonna [pic in store window  left?]
26. Red hot chili peppers
27. Twisted sister  the contortionists in the middle.
28. Eagles?