Mark:
PALINDROMES FROM 0 TO 9999999
Consider the seven-digit number, ABCDEFG, where leading zeros are allowed. All palindromes are determined by the values of A, B, C, and D (E=C, F=B, and G=A). All possible values for A, B, C, and D lead to palindromes (some with leading zeros). The palindromes with leading zeros can be converted to shorter palindromes by removing the leading (and trailing) zeros. This will generate all palindromes of odd length up to seven.
Even length palindromes can be generated the same way with the six-digit number, ABCDEF. Here, A, B, and C determine the palindrome.
Poor Mark, he must be suffering from jet lag. Although his logic cannot be questioned, what the hell is he talking about? A,b,c diddle d. I ask; can anyone translate? I did not spend three weeks learning English just to translate Chicago speak.
What he meant to say was:
There are 10,999 palindromes between zero and ten million. It's best to count them according to the number of digits they have.
There are 10 one-digit palindromes.
There are 9 two-digit palindromes (11, 22, 33, . . . , 99).
There are 90 three-digit palindromes, since there are 9 choices for the outside digits (we can't use zero there!) and 10 choices for the inside digit.
There are 90 four-digit palindromes, since again there are 9 choices for the outside digits, and 10 choices for the inside digits. Note that both of the two inside digits must be the same.
There are 900 five-digit palindromes, since there are 9 choices for the outside digits, 10 choices for the next-to-outside digits, and 10 choices for the middle digit.
There are 900 six-digit palindromes, since there are 9 choices for the outside digits, 10 choices for the next-to-outside digits, and 10 choices for the two middle digits (which must both be the same).
There are 9000 seven-digit palindromes for reasons similar to those described above.
The only eight-digit number to consider is 10,000,000, which is not a palindrome. So, the number of palindromes between 0 and 10,000,000 is
10 + 9 + 90 + 90 + 900 + 900 + 9,000 = 10,999
What could be easier?
It was at this point that I found I had missed Mark's last line,
which read,
"There are 9,999 possible odd length palindromes and 999 even length palindromes, not counting zero. Therefore, there are 10,999 palindromes less than 10,000,000."
On which we all agree.
You are correct, my Dr says I am a psycho. However, "And she was quite the maniac. Neither of us made it to our conventions, but we did dress up in each other's clothes."
Whoa! That's way too much information for me to handle. Although, I suspect you managed ok.
At this point I would like to thank Rap for standing in at very short notice, in an effort to keep this train wreck going.
Having taken care of the Ugly, the Good and the Bad are dividing a bounty of 2000 silver dollars as follows. First, the Good divides the coins into two piles with at least two coins in each pile. Then the Bad divides each of those two piles into two (non-empty) piles and takes the largest and the smallest of the four piles, leaving the other two piles for the Good.
What is the maximum share the Good can count on no matter how smart - or greedy - the Bad is
a) Fred Flintstone and Barney Rubble in turn take pebbles from a pile of 2000 pebbles. They can take 1, 7 or 13 pebbles at a time. The player who takes the last pebble wins.
Find a strategy that allows Fred or Barney to win regardless of how the other one may play
Oh, Fred goes first.
(b) What is the winning strategy if the players are allowed to take 1, 2, 7 or 13 pebbles at a time
