I just took it as a challenge to myself to see if I could get the entire number by reasoning alone without resort to trial and error. I actually did get the number by trial and error first, and then figured out the reasoning after the fact.

Its often much easier to solve a problem when you know the answer. Though I did the same thing; tried to figure out after my brute-force approach to the last digit what a more analytical method of figuring it could be. Turned out to be just a couple of easy steps, I was just lazy and didn’t try when I should have.

]]>Just a minor quibble. In the first line on the second problem, I was temporarily confused. Your notation has white’s second move as 2. Rxa4+. You really should have written this as 2. Rcxa4+ since the rook on the a file could take the a4 pawn and give check. Clearly that move doesn’t lead to stalemate since black would be able to capture that rook on a7.

Like I said, I minor thing, but I was a bit confused on first reading. Probably that’s on me, but adding the “c” to the notation would help.

]]>…and a well-deserved spanking:

http://www.evolutionnews.org/2014/09/metaspriggina_r089761.html

]]>For which n is there a n-digit number that reverses when you multiply by 4?

]]>Actually, it is possible to get all the digits by mathematical reasoning without resort to any trial and error. Of course, in the interest of allowing Jason’s students (and anyone else who might want to come up with it on their own), I will not post any more until next week after solutions are due.

]]>Oh, don’t get me wrong, I still enjoyed it. But thus far I found Week 1 (with the extended number sequence – I never did get 31 until it was posted) the hardest

Week 1 wasn’t hard as written. What we struggled with was the numbers higher than 10 – but the students didn’t have to solve for those.

]]>My 2-4 don’t resemble yours at all, either. My #1 is the same. After that, it goes:

2. (9+1)*(20-15-4)

3. (21+3)*(6-5)/2

4. (19+5-23)^(7+11)

5. (25-22)*((21-20)^18)

A basic rule of thumb for these seems to be: get the answer using a subset. Then try and make the remaining numbers into a 1 or 1^N, where N can be the sum of any numbers you haven’ figured out what to do with yet. ]]>

The Schwarzenegger palindrome was amusing.

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