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Posts posted by Pickett


I think this works (basically paint the whole plane one color...then paint 2 intersecting lines another color...and then paint the intersection point of those two lines a third color):
(1a) solved
You could use as many blue lines as you want (if all of them are intersecting in black point and if some red points remain). One blue line seems most elegant to me. Just wandering, why you chose to use two blue lines.
good question...I guess I just immediately went to quadrants when talking about "the plane"...


the clouds should be above the blue in the sky?
Why do the clouds have to be above the blue? Could just be that a storm is brewing on the horizon...
So, when I said you can't see the river on the other side of the bridge, I should have gone with the opposite: you can't see the other side of the bridge under it. The bridge is relatively narrow (~10 feet or so)...and yet, when you look under it, it's completely black...You can't see the sunlight on the other side of it. With a bridge that narrow, you would DEFINITELY be able to see the other side or at least some sunlight showing through.

I just read the "Normal People Option" and I definitely think the answer is Shadows

I actually have just been watching the new Dr. Who...and at FIRST I was thinking the answer was "Angels" because of the blinking line...However, after finishing reading the "Whovian Option", I believe the answer is "Shadows", or more specifically the "
Vashta Nerada"... 1

Is that all?
I didn't see the river on the other side of the bridge...however, it COULD just take a hard right and remain hidden behind the bridge...so, I don't think that's it...

...would I cross that bridge...the keystone is upside down and therefore doesn't really provide much structural support. The arch will just collapse in on itself!

The more attractive people would have more people behind their backs

You typically play assuming that the dealer has a 10 card down...meaning you have to assume they have 16. You have either 17 or 7...this means that by getting another card you have NO way of going over 21.
You didn't mention what the dealer's rules are, but in most casinos they HAVE to hit if they have under 17, and HAVE to stay if they have 17 or more...if that is the case, then you should double down, because you will double your bet and get a card (can't go over 21)...and then the deal HAS to hit, which means they will most likely go over 21. In the end with this approach you will win double your original bet.
In practice, however...this never happens...I always seem to lose in these "simple" scenarios

Short answer: SUNDAY.
Longer answer: Obviously it depends on your time constraint...but I assume you just mean "over all of time (after the Gregorian Calendar was established)"
Sunday, Tuesday, and Friday will start the year more often than any other days (and equally so). Then Wednesday and Thursday...and finally Monday and Saturday.
For example: over 10,000 years:
Sunday  1450
Monday  1400
Tuesday  1450
Wednesday  1425
Thursday  1425
Friday  1450
Saturday  1400and over 100,000 years:
Sunday  14500
Monday  14000
Tuesday  14500
Wednesday  14250
Thursday  14250
Friday  14500
Saturday  14000
Obviously this is due to the 365 days per year means every year the starting day shifts by 1...except every 4 years it shifts by 2...except every 100 years it shifts back one...and then every 400 it moves ahead one...I could go into the math to really prove everything out...but I'll leave that for someone else...Neat puzzle...very simple question that was fun to solve.

GOOD NEWS: If the clock hands only move once per minute, then this clock is good enough! There are no completely ambiguous times!
BAD NEWS: If the clock hands move every second (which is more likely), then from my calculations there are 66 pairs of times that you can not tell which it is...here are a few of them (I won't list them all):
04:31:50 06:22:3502:56:10 11:14:4003:41:30 08:18:2509:54:05 10:49:3001:55:45 11:09:3503:31:25 06:17:3512:55:20 11:04:35...
I didn't go as far to try to calculate a CONTINUOUSLY moving hand, but that would open up a whole new set of possibilities...Now, while I agree that the old man is THEORETICALLY correct that it's pretty straightforward to tell what time it is using his clock, I, for one, will not be purchasing one of these...

Can you put six points on the plane, so that the distance between any two of them is an integer, and no three are collinear?
These 6 coordinates work:
(7225, 0)(7225, 0)(4025, 6000)(4025, 6000)(2023, 6936)(2023, 6936) 
n is the number of squares desired
m is the number of lines needed
n 2, 3, 4 5, 6, 7, 8, 9, 10 11, 12, 13
m 5, 6, 7 6, 7, 8, 7, 8, 9 8, 9, 10
5, 6, 7
6, 7, 8
7, 8, 9
8, 9, 10
any hypothesis as to why this is happening?
and can anyone confirm pickett's 14 and 15 calculation?
The pattern definitely doesn't continue, because you can certainly do 14 squares in 8 lines...just make a 3x3 box:
1x1=9, 2x2=4, 3x3=1...14 squares, 8 lines...and then just add one more line 3 units away from the 3x3 to create an additional 3x3 and you have 15 squares with 9 lines:
However, I went ahead and did attempted the next few:
14=8
15=9
16=10
17=9
18=10
19=11
20=10
21=11
22=12
23=11
24=12
25=13
26=10
27=11
28=12
...skip one
30=10
...skip a few
40=11
...skip a few
50=12
...skip a couple
55=12
...skip a few more
70=13
...skip a lot
91=14
...skip lots and lots
140=16
...and why not...if you want 29370 squares, you can do it with just 90 lines!!

25 lbs
nice work! Please elaborate on your answer.
Basic system of equations with 3 variables...from the statements:
Let b = baby's weight, d = dog's weight, w = woman's weight...
if the mother weighs 100 pounds more than the combined weight of the baby and the dog
EQUATION 1: b + d + 100 = w
the dog weighs 60 percent less than the baby
EQUATION 2: d = 0.4*b
and then from the picture, we know
EQUATION 3: b + d + w = 170
Then just solve the system
Substitute equation 2 into equations 1 and 3:
1) b + 0.4b + 100 = w and 3) b + 0.4b + w = 170
1) 1.4b + 100 = w and 3) 1.4b + w = 170
Then substitute equation 1 into equation 3:
1.4b + 1.4b + 100 = 170
Solve for b:
2.8b = 70
b = 25

I don't think it works. Line is infinite, whereas you used line segments. In my thinking above lines create 8 square.
I actually still stand by all of the numbers I originally posted...
1: 4
2: 5
3: 6
4: 7
5: 6
6: 7
7: 8
8: 7
9: 8
10: 9
11: 8
12: 9
13: 10
14: 8
15: 9
There may be better, but I was at least able to get each number 115 using the number of (infinite) lines as I said originally...That's kind of neat, considering the square constructions had to be completely different than what I doing using line segments...but still got the same numbers...

How did you get 6:7?
__ __ __
______
____
1x1: 5
2x2: 1
7 lines, 6 squares
YAY ASCII ART!...
I don't think it works. Line is infinite, whereas you used line segments. In my thinking above lines create 8 square.
I agree though, Pickett's answer appears to give 8 squares.
Ahh, didn't realize the OP meant infinite lines, not line segments...that definitely makes the puzzle more fun/interesting

sugar
Sugar
cube (hence the me x me x me part...) 

How did you get 6:7?
__ __ __
______
____
1x1: 5
2x2: 1
7 lines, 6 squares
YAY ASCII ART!...

Found/Fond
Resign/Reign

I actually also don't even see any 5, 6, or 7 move solutions (although with 7 moves, I only validated with n < 10...since my algorithm is extremely inefficient...)...so I'm starting to think that the ONLY solutions to this are the 4 move solutions presented.

I proved programmatically that for n < 5000, there are no 3 move solutions...for n < 50000 there are no 2 move solutions...and for n < 500000 there are no 1 move solutions...and for n < 500, the only 4 move solutions are:
1,1,1,5
1,1,5,1
1,5,1,1
5,1,1,1
Plus, once you get into higher numbers you can prove that the right hand side of the scale is going to simply grow faster and faster (since it's multiplication and the left side is addition)...So there really is no need to test with n higher than what I have already done...

Just a little bump because I'm curious if anyone else has looked at this one and if my solutions above are indeed optimal or not...

Nice.There are MANY(at least 51) that meet these criteria...here are all 51 with side lengths under 1000...10x87 1x29011x48 1x17612x35 1x14015x22 1x11020x174 2x58021x122 2x42722x96 2x35224x70 2x28026x57 2x24730x44 2x22030x261 3x87031x42 2x21733x144 3x52836x105 3x42040x81 3x36042x244 4x85444x192 4x70445x66 3x33048x140 4x56049x132 4x53952x114 4x49453x54 3x31855x240 5x88058x195 5x75460x88 4x44060x175 5x70062x84 4x43468x75 4x42570x123 5x57471x120 5x56872x210 6x84075x110 5x55078x171 6x74180x162 6x72081x158 6x71184x95 5x53284x245 7x98089x210 7x89090x132 6x66093x126 6x651102x161 7x782104x228 8x988105x154 7x770106x108 6x636111x200 8x925112x141 7x752120x176 8x880124x168 8x868135x198 9x990136x150 8x850159x162 9x954...
Curious about your algorithm.
Mine ran forever with n=100, (and came up empty.)
I tried direct solutions also, restricting the first to be square and the second to be 1xm.
Came up empty again.
It's unbelievably inefficient and very brute force, but it worked like a charm:
public class Dummy { public static void main(String... args) { for (int a = 1; a < 1000; a++) { for (int b = a; b < 1000; b++) { for (int c = 1; c < 1000; c++) { for (int d = c; d < 1000; d++) { if ((a * b) == 3*(c * d) && 3 * (2 * a + 2 * b) == (2 * c + 2 * d)) { System.out.println(a + "x" + b + " " + c + "x" + d); } } } } } } }
Stable Job
in New Logic/Math Puzzles
Posted · Edited by Pickett · Report reply
On an unrelated note...why did someone vote this as only 2 stars?? I think this was a really good puzzle and it was fun to solve...