Wikipedia:Reference desk/Archives/Mathematics/2009 August 1

= August 1 =

computing modulo
I want to compute the remainder when $$2^{2^{17}}+1$$ is divided by 19. The hint in the book that I have says first compute $$2^{17}$$ mod 18. How can I do that please? I can't apply Fermat's theorem or Euler's theorem directly, yet as these are the only results proved in the book upto this point I am pretty sure that I am expected to use them only. Thanks.--Shahab (talk) 14:45, 1 August 2009 (UTC)
 * Try computing the smaller powers of 2 mod 18 first. Michael Slone (talk) 16:30, 1 August 2009 (UTC)
 * ....and observe the pattern that they follow. Michael Hardy (talk) 16:46, 1 August 2009 (UTC)
 * OK. 2^17=14. What should I do now. I want to get a result mod 19 somehow.--Shahab (talk) 18:35, 1 August 2009 (UTC)
 * Working mod 19, 2^14=2^-4=16^-1, so you could compute the inverse of 16. Or compute the inverse of 2 and square it twice. Or just divide 16384 by 19 and find the remainder. Algebraist 20:21, 1 August 2009 (UTC)
 * You now have enough information to apply Fermat's Little Theorem.84.202.41.200 (talk) 22:55, 3 August 2009 (UTC)

Working together word problem
The problem reads: Abbie paints twice as fast as Beth and three times as fast as Cathie. If it takes them 60 min to paint a living room with all three working together, how long would it take Abbie if she works alone?

Here is what I have: $$A=2B,\,\!\,A=3C$$ $$\frac{A}{2} + \frac{A}{3} + A = 60$$ $$\frac{11A}{6}=60$$ $$A=\frac{360}{11}$$ Which I know is incorrect because I know the answer is actually 1 hour 50 minutes. Where have I gone wrong, and how do I do it correctly? TIA, Ζρς ιβ' ¡hábleme! 17:02, 1 August 2009 (UTC)

In your first equation you treat A, B, and C as the rates at which they work, and in the second as the times. They can't be both: if you make the rate bigger, you make the time smaller.

The rate is inversely proportional to the time. If Abby can do the job in a hours, she can do 1/a of the job in 1 hour. Beth would then take twice as long: 2a hours; and Cathie would take 3a hours. How much they could do in one hour would be 1/a, 1/(2a), and 1/(3a) for Abby, Beth, and Cathie respectively. So you have 1/a + 1/(2a) + 1/(3a) = 1. That gives a = 11/6 hour = 1 hour and 50 minutes. Michael Hardy (talk) 17:46, 1 August 2009 (UTC)

If you take A to be the rate, then the only thing wrong is the RHS of the second equation. The rate of the three of them working together is indeed A + A/2 + A/3, but the total rate given in the problem is 1 living room per 60 min, which is 1/60 instead of 60 (assuming you're working in minutes). It might be helpful to include the units in your equation. Personally I find that helps avoid mistakes although your mileage may vary. Rckrone (talk) 22:28, 1 August 2009 (UTC)

Two cubes that share three corners
Is it possible for two cubes of the same size to share exactly three corners? Neon Merlin  23:59, 1 August 2009 (UTC)


 * Yes. Let the vertices of a cube be the eight points in {0, 1}3.  Consider the three points
 * (1, 0, 0)
 * (0, 1, 0)
 * (0, 0, 1)
 * Consider the plane in which they lie to be a mirror. The mirror-image of the cube described here is another cube that shares those same three vertices.
 * (I wrote a longer reply and it was destroyed by an edit conflict when someone put in the August 2nd heading.) Michael Hardy (talk) 00:11, 2 August 2009 (UTC
 * You know, you can rescue text lost to an edit conflict - your text is in the textarea at the bottom, you can copy and paste it from there to the top box and submit it. Alternatively, you can copy it from the diff, but that requires a little tidying up afterwards (it is my preferred method, though, since you don't have to search for your text). --Tango (talk) 15:38, 2 August 2009 (UTC)
 * You know, you can rescue text lost to an edit conflict - your text is in the textarea at the bottom, you can copy and paste it from there to the top box and submit it. Alternatively, you can copy it from the diff, but that requires a little tidying up afterwards (it is my preferred method, though, since you don't have to search for your text). --Tango (talk) 15:38, 2 August 2009 (UTC)

OK, more long-windedly: The eight corners of one cube are these:
 * (0, 0, 0)
 * (1, 0, 0)
 * (0, 1, 0)
 * (0, 0, 1)
 * (1, 1, 0)
 * (1, 0, 1)
 * (0, 1, 1)
 * (1, 1, 1)
 * (0, 1, 1)
 * (1, 1, 1)
 * (1, 1, 1)

The eight corners of the other cube are these:
 * (2/3, 2/3, 2/3)
 * (1, 0, 0)
 * (0, 1, 0)
 * (0, 0, 1)
 * (2/3, 2/3, &minus;1/3)
 * (2/3, &minus;1/3, 2/3)
 * (&minus;1/3, 2/3, 2/3)
 * (&minus;1/3, &minus;1/3, &minus;1/3)
 * (2/3, &minus;1/3, 2/3)
 * (&minus;1/3, 2/3, 2/3)
 * (&minus;1/3, &minus;1/3, &minus;1/3)
 * (&minus;1/3, &minus;1/3, &minus;1/3)

Michael Hardy (talk) 03:34, 2 August 2009 (UTC) The three corners of the white triangle are the three vertices that the two cubes share. One corner of each cube&mdash;the one that is "inside" the white triangle&mdash;is within the interior of the other cube. Michael Hardy (talk) 03:42, 2 August 2009 (UTC)
 * hallo! I took the liberty of kicking up the cube... pma (talk) 19:40, 2 August 2009 (UTC)