Just a few days ago, I received a lovely surprise message from Bob Mitchell, a Scottish chess friend whom I haven’t seen for very many years. I remembered Bob immediately, as he was always friendly and funny as well as being a talented, attacking chess player. The descriptive words FUN BATTLE GOER come quickly to mind!
Here’s a sneaky word puzzle in Bob’s honour: rearrange the letters of
FUN BATTLE GOER
to make a proper 13-letter English word which also describes Bob!
You’ll always happily remember the answer because it’s
Bob’s opponent in the following beautiful game from 1999 is Graeme Nolan, who is actually a rather good player, but (as you’re about to witness) Bob’s play was highly energetic and inspired.
White: R.Mitchell; Black: G.Nolan; 1999 Marymass Open
1 e4 e6 2 d4 d5 3 exd5 exd5 4 Nf3 Bd6 5 c4 c6 (5…Ne7?? 6 c5 would be embarrassing for the black bishop then trapped on d6) 6 Nc3 Ne7 7 Bd3 Bg4 8 0-0 Nd7 9 cxd5 cxd5 10 Re1 Nb6 (Black understands that White was threatening Nxd5, and he also knows that 10…0-0 would allow 11 Bxh7+! in view of 11…Kxh7 12 Ng5+ Kg8 13 Qxg4) 11 Nb5 (a good alternative is 11 a4, threatening a5) 11…Bb8?! (11…Bb4 is better) 12 b3 a6 13 Ba3 Nc8 14 Rc1! (threatening Rxc8! to undermine the support of the pinned knight on e7) 14…Be6 15 Ng5! (after that further increase in the attacking power of White’s pieces, Black’s position is indefensible) 15…axb5 16 Bxb5+ Kf8 17 Rxe6 (17 Qf3 wins too, as 17…Kg8 is met by 18 Rxe6) 17…Rxa3 18 Qf3 f5
(I cannot resist noting another beautiful tactical point: if the rook on c1 were instead on e1, then 19 Qxf5+! Nxf5 20 Re8+ Qxe8 21 Rxe8# would make a stunning checkmate!) 19 Qh5 g6 (19…Nd6 doesn’t help, as White is ready with 20 Rxd6! Bxd6 21 Qf7#) 20 Qh6+ Kg8 21 Rxc8! Qxc8 (or 21…Nxc8 22 Rxg6+ hxg6 23 Qxg6+ Kf8 24 Qf7#) 22 Rxe7 Qc1+ 23 Bf1, and Black played just a few more moves before resigning, as his king is defenceless against White’s overwhelming threats. 1-0.
Fun Numbers Brainteaser
is a four-digit whole number.
Your fun challenge is to figure out the largest possible value for the digits AB such that the four-digit number 1AB1 is a multiple of AB.
(Note: In this puzzle, we are not counting AB as A x B, but rather as a two-digit number AB.)
A solution is given at the end of the article.
Fun Surprise for Valentines’ Day!
- Write down your favourite three-digit whole number. Let’s call it XYZ. Some or all of the digits can be equal, if you like, but X should not be zero.
- Now form the six-digit whole number XYZXYZ.
- Divide by 91, given that this is Blog Post #91.
- Divide by your chosen three-digit number, XYZ.
- Add on my personal favourite number, 3.
I wish you a really happy Valentines’ Day now on Sunday 14 February!
With very best wishes as always,
Paul Motwani xxx
By considering the place-values of the digits in 1AB1, we can realise that the four-digit number equals 1001 + 10 x AB. If we want that to be a multiple of AB, then (since 10 x AB is clearly a multiple of AB) we’ll need 1001 to also be a multiple of AB.
Now, 1001=7 x 11 x 13. Its largest two-digit factor is therefore 7 x 13 = 91.
So, the maximum possible value of AB which meets the requirements is:
AB = 91.
1911 = 21 x 91.