Nötknäpparen. (The Nutcracker.)

Nötknäpparen was a Swedish crossword-paper, my last issue is from 1979 when, I think, the paper died at an age of 26 year. I bought my first Nötknäpparen in september 1964, I was already familiar with it, my parents loved crosswords. My favorite column was:

Knepiga Klubben (The Tricky Club)

where the readers could try solving, or publish, a various of tricky mathematical problems. In the summer 1967 the new chairman of the club, Lars Bertil Owe (alias Lasse Knepson) took the problems to a new hight, and the club-members (readers) even cracked some old nuts. Another column was Nötknäpparnas Egen Sida (The Nutcrackers Own Page) with easy math- and word-puzzle ideas from the readers (nutcrackers). The club slowly died 1974-1975, after the owners (Olle Cronsjö) death, when the new owner slowly gave the paper a new direction and later a new layout.

The Cracked Nuts.

October 1969. Placing Halfpennies.

Problem no. 306 in Curious and Interesting Puzzles by David Wells. My quote is from this book:
Here is an interesting little puzzle sugested to me by Mr W. T. Whyte. Mark off on a sheet of paper a rectangular space 5 by 3 inches, and then find the greatest number of halfpennies that can be placed within the enclosure under the following conditions. A halfpenny is exactly an inch in diameter. Place your first halfpenny where you like, then place your second coin at exactly the distance of an inch from the first, the third an inch distance from the second, and so on. No halfpenny may touch another halfpenny or cross the boundary. ...some comment on an illustration.... How many can the reader place? (H. E. Dudeney, 1917 no. 429)
(Dudeney's solution was 13 halfpennies.)
(You can find Dudeney's book here: Free eBooks - Project Gutenberg)
7 members in the Club had the solution 14 halfpennies. My solution:
We call the radius of the halfpenny 1 and the corners of the rectangle (0,0), (10,0), (10,6) and (0,6). Then place halfpenny 1 in (7,3+x), 2 in (3,3+x), 3 in (y,5-x), 4 in (7,1), 5 in (3,1), 6 in (z,w), 7 in (1,1+x), 8 in (5,1+x), 9 in (9,1+x), 10 in (2+y,5), 11 in (y-2,5), 12 in (1,3+2x), 13 in (5,3+2x) and 14 in (9,3+2x). The constellations of halfpennies (11,13,8), (3,1,4) and (10,14,9) are similar. Call the distance, between halfpenny 1 and 3, d, we will check for overlap. Then we have:
Likewise 3 and 4 give: (7-y)**2+(4-x)**2=16
Those equations give us : 4+3x**2=d**2, than d is greater than 2.
Coins 2 and 3 give: (y-3)**2+(2-2x)**2=16
and coins 3 and 4 give: (7-y)**2+(4-x)=16
Those last two equations give us: (20-3*x**2)**2+256*(1-x)**2=1024
x=0.03 make the left part=1024.7192973 and
x=0.04 make the left part=1019.660823 then
x greater than 0.03 and less then 0.04. This will give us all the coordinates.

February 1970. Bookprices.

A buyer could not decide which six out of eight books he should buy. The prices were in whole dollars not one less then 2 dollars. The price of every six books adds up to a different number. Give the smallest amout of money he had to pay when he at last decided to buy all eight books. (Problem 20 in Unterhaltsame Mathematik by Roland Sprauge)
Sprauge gave the answer: 2+3+4+6+10+15+20+30=90.
But 16 members had the answer: 2+3+4+6+10+15+21+26=88.
Later I found the same answer in problem C9 from Richard K, Guy's book Unsolved Problem in Number Theory.

May 1970. Cut twelve pentominoes out of plywood.

A problem from Polyminoes by Solomon W. Golomb, I quote the book:
A man wishes to construct the twelve pentominoes (a shape that cover five squares) out of plywood. His saw will not cut around corners. What is the smallest plywood rectangle from which he can cut all twelve pentominoes? (The U-shaped pentomino will require special effort. Assume it will be cut as a 2*3 hexomino, a hexomino is a shape that cover six squares, and finished it later.)
The answer in the book was 6*13.
But eight members had the answer: 5*15, Mr Adolf Ahl: 7*(8.5+1.5*sqrt(2)) and Mr Fai Lötberg: 8*9. His solution:
LLFOIVVVZ  and free them in that order
NFOPIYZUU  and free them in that order
NNPPIYYUO  5. O is waste
ONPPIYOUU  6. observe the diagonal cut in 3

A 5*15 from my hand:

More pentominoes.

September 1972. Pentominoes and single holes.

The problem came from Martin Gardners column in Scientific American, after an idea from Stephen Barr.
Place the twelve pentominoes so there are as many single holes as possible.

a) No holes connected in their corners.

b) Corners connection allowed.

The best solver was Mr Christer Lindstedt, 13 and 18 holes :

April 1973. Pentominoes used as a fence.

The problem was made after an idea from Christer Lindstedt.
Cut an 8x8 square into the twelve pentominoes and a 2x2 square. Use all the 13 pieces as a fence around an as big as possible area. The pieces should be connected in the domino way.
The 13 pieces as a square:

The best solution, with 143 fenced squares, was from Mr Ole Björkqvist and Mr Sune Backan.

Made by Sven Egevad
Epost: sven.egevad@telia.com