Saturday, March 27, 2010


Blog post 56:

I don't have much to post today. But I like these last two pictures I made (made yesterday and the day before). So, I definitely had to post something today to show you them.


This Fraying Of Palindromes


Let's go over the result of my 3 most recent polls (which are now closed).

For favorite fundamental force of nature, gravity wins in a landslide, winning all 4 votes the poll received.
(I am not surprised by this result.)

For favorite 2-D shape, hexagon received the most votes, with 2, while rhombus, parallelogram, and octagon each received one vote.
(I am surprised that no one picked circle.)

For favorite 1-digit number, 3 and 8 tied for 1st place, with 2 votes each. 7 received 1 vote.
(I would have voted for 0 or 1 if I had voted.)

Look for my next poll, whenever and whatever that will be!


Update: Regarding my confession over the expired cottage cheese (See my last post).

I went back to the 7-11 two days later, and I did indeed then tell the clerk/manager about the expired cottage cheese.
In the meantime, however, someone had bought one of the expired tubs.

Hope they didn't get sick!


Finally, a puzzle:
(I haven't had one of those in a while.)

Here is an unoriginal puzzle (stealing mostly from Soduku).
The puzzle itself is easy to solve, but it was enjoyable for me to invent and to prove that there is only one solution.

The 10 variables (m(0),m(1),m(2),...,m(9)) when their values are written out form a permutation of the integers (0,1,2,3,4,5,6,7,8,9).
Each individual 1-digit integer equals exactly one m-variable. But which one?

What is the permutation?

As I said, there is only one solution, unless I made an error.

Here are the conditions:

A) m(0) + m(1) + m(9) = a composite.

B) m(6) = a prime.

C) 1 = m(p), for p = some prime.

D) m(2) + m(3) = m(4), and m(2) > m(3).

E) m(m(1)) = 4.

F) m(4) = a square integer.

G) There are exactly two (no more, no fewer) integers n such that m(n) = n.

H) If you take the solution to this puzzle, the values of (m(0)m(1)m(2)...m(9)), and write the values out from left to right, concatenating the digits to form one huge number (all in base-10, of course), then m(8) = the second smallest prime that divides this large number.

For you at home who don't know what a composite or a prime are, all you have to know for this puzzle is that the sequence of primes begins: 2,3,5,7,11,13,17,19,23,29,...
And the sequence of composites begins: 4,6,8,9,10,12,14,15,16,18,...

Answer later posted in the comments section of this particular blog-post.

Leroy Quet

1 comment:

Amorphous Trapezoid said...

The answer:


Or for those of you preferring commas between your numbers:

4, 0, 8, 1, 9, 5, 2, 7, 3, 6