- About Us
- Chess Improvement
- Chess Puzzles
- Chess Research
- College Chess
- General News
- Home
- Major Tournaments
- News
- Polgar Events
- Privacy Policy
- Scholastic Chess
- SPICE / Webster
- Susan’s Personal Blog
- Track your order
- USA Chess
- Videos
- Women’s Chess
- Contact Us
- Daily News
- My Account
- Terms & Conditions
- Privacy Policy
1. e3…Nf6
2. e4…Nxe4
3. Ne2…Nxd2
4. N(e2)c3…Nxb1
5.Nxb1
By Craig & Jim S.
Please Susan…
Resolve the issues with challenge #3 before moving on to challenges #4 and #5.
Regarding challenge #3, you should post either a real solution or a correction. Because the solution that comes up in the comments is just bogus, and it really looks like the problem is flawed.
Building on the ideas used in the second Benko puzzle, it’s not too hard to come up with this:
1- e3; Nf6
2- e4; Nxe4
3-Ne2; Nxd2
4- Nec3; Nxb1
5- Nxb1
1.Nf3 Nf6
2.e4 Nxe4
3.Ne2 Nxd2
4.Nec3 Nxb1
5.Nxb1
I like:
1. e3 Nf6
2. e4 Nxe4
3. Ne2 Nxd2
4. Nec3 Nxb1
5. Nxb1
Brad Hoehne
Is it…
1. e3 Nf6
2. e4 Nxe4
3. Ne2 Nxd7
4. Nec3 Nxb1
5. Nxb1
I think.
—
#4 is as easy as #3:
1. e3 Nf6
2. e4 Nxe4
3. Ne2 Nxd2
4. Nc3 Nxb1
5. Nxb1
The key was to notice that the last move by black must be a knight capture.
1. e3 Nf6
2. e4 NXe4
3. Ne2 NXd2
4. Nc4 NXb2
5. NXb2
This is a cute one.
1. e3 Nf6
2. e4 Nxe4
3. Ne2 Nxd2
4. Nec3 Nxb1
5. Nxb1
which replicates the diagram (and is legal, unlike challenge #3!).
1. e3! Nf6
2. e4 Nxe4
3. Ne2 Nxd2
4. Nec3 Nxb1
5. Nxb1
1. e3 Nf6
2. e4 Nxe4
3. Ne2 Nxd2
4. Nec3 Nxb1
5. Nxb1
1. e3 Nf6
2. e4 Nxe4
3. Ne2 Nxf2
4. Nec3 Nxb1
5. Nxb1 and done!
And I agree with “Anonymous”. Please post the solution/correction for Challenge #3. I can’t believe that we can do 1, 2 and 4 but 3 evades all of us.
1. e3 Nf6
2. e4 Nxe4
3. Ne2 Nxd2
4. Nc3 Nxb1
5. Nxb1
this is solvable! yuppie!
Susan…
Challenge #3 has Nf6+ followed by Nb1 is illegal…
Please post a solution for that first…
This solution is very easy..
e3 Nf6
e4 Ne4
Ne2 Nd2
Nec3 Nb1
Nb1
1. e3 Nf6
2. e4 Ne4
3. Ne2 Nd2
4. Nec3 Nb1
5. Nb1
OK, this one can be done.
1.e3 Nf6
2.e4 Nxe4
3.Ne2 Nxd2
4.Nec3 Nxb1
5.Nxb1
1.e3 Nf6 2.e4 Nxe4 3.Ne2 Nxd2 4.Nec3 Nxb1 5.Nxb1
i think i found a way to do it:
1. e3 Nf6
2. e4 Nxe4
3. Ne2 Nxd2
4. Nc3 Nxb1
5. Nxb1
greets and merry x-mas, jan
1.e3 Nf6
2.e4 Ne4
3.Ne2 Nd2
4.Nec3 Nb1
5.Nb1
yippie !!
1.e3 Nf6
2.e4 Ne4
3.Ne2 Nd2
4.Nec3 Nb1
5.Nb1
1.d4 Nf6 2. d5 Nxd5 3. e3 Nxe3 4. Bishop any Nf1 5. Bxf1
1- e3 nf6
2- e4 ne4
3- ne2 nd2
4- ne2-c3 nb1
5- nb1
A.Wailer
You people are all missing the forest on #3 by looking at the trees. If you consider question (b) first, the solution becomes quite obvious.
1. Nc3 Nf6
2. Nd5 Ne4
3. c3 N:c3
4. N:e7 Nb1
5. Ng8
It is black’s e-pawn that must be missing. This is the only possible solution to question (b), since both knights must move 4 times. Therefore there is only one pawn move, and it must be c2-c3 to allow it to be captured. But that’s white’s 5th move, so all other moves must be N moves and white must be the one to capture a pawn. The pawn at e7 is the only one that can be captured and still allow White’s QN to get to g8 in 4 moves.
1. e3 Nf6
2. e4 Nxe4
3. Ne2 Nxd2
4. Nec3 Nxb1
5. Nxb1
-Ridhnar
1. e3 Nf6
2. e4 Nxe4
3. Ne2 Nxd2
4. Nec3 Nxb1
5. Nxb1
the #3 question was edited
it is legal now