Interesting math problems

[lendav_rott]

"Interesting" math problems

Since I have naught to do right now and the weather is not great whatsoever, I'm going through a Math problem book I found covered in dust sitting on the shelf at my folks' - it's from 1960 entitled (if word for word translation) "Mathematical Sharpness" by B. Kordemsky originally in Russian, but I do my best to translate for yous ^^

Interesting doesn't necessarely mean difficult, but I looked ahead a bit and this book will end up in complete fireworks in terms of difficulty. Contains 369 different puzzles and I will be working my way through this book for the days/weeks to come and will post, after I've solved something, the ones I like/had trouble with or both, who knows.

I'll number the puzzles the way they are given in the book, perhaps someone recognizes the book and the puzzles :)

6)
http://www.upload.ee/image/3584987/Puzzle6.jpg
A gardener, represented as * is attending to his apple trees, which are represented by the dots.
The gardener cannot move diagonally, may not enter the red squares and may not walk on already treaded path. Show a way for the gardener to attend to Every tree and end up where he started.

I like this one since I used to be a hardcore nokia 3310 snake player and I kind of saw the path almost instantly, made me smile :)

8) Here's one that reads so funny to me and if you're a visual thinker, it might make you go haywire at first:

How many cats are in a quadrangular room if in every corner of the room there's a cat, every cat in the room is facing 3 cats and there's a cat sitting on every cat's tail.

9) This one is tricky, so simple, but yet so difficult to see :D

There are two pencils red and green held together so that their bottom parts are in the same position ( you hold two pencils together and set them on a horizontal plane to "even" them out). The red pencil's bottom part is covered in paint along 1 cm. While holding the pencils still tightly together, you begin to move the red pencil 1 cm down and then 1 cm back up, down 1 cm, up 1 cm - the green pencil stays fixed. You complete the up-down cycle 10 times hence you make 20 moves total. Once finished, what is the length of the stain of paint that is covering the red pencil if the paint does not dry or fade?

 

[arildno]

This could be fun!

 

[lendav_rott]

22)
We have the numbers 1 2 3 4 5 6 7 8 9, each number can only be used once.
One of each peaks of a triangle contains a number from the above selection. Distribute the numbers between the peaks and the sides of the triangle such that the sum of each side of the triangle will be 20 (a sum of the side being peak+peak2+theside in between). How many different solutions are there?

This one was thoroughly enjoyable, since I worked on it almost all day and trying to determine all of the conditions for it to work and devise a universal formula for any such assignment (If there is one already, I am not interested, I want to invent the wheel for myself ^^ )

 

[mfb]

Somehow that looks like a lot of casework. I get 16 cases with a 5 in one corner, not counting rotations and reflections.

 

[lendav_rott]

I began disecting it like this:
If A, B, C are the peaks and x, y, z the separate sums on the sides then:
A+ B +x = 20
A+ C +y = 20
B+ C +z = 20
so add that all together
let x+y+z = D
2(A+B+C) + D = 60
D has to be an even figure, because 2(A+B+C) is even, regardless if the sum of the peaks is odd or even.
As D is even, the numbers left after we pick the peaks will have to contain an even number of odd numbers otherwise the leftover sum will be odd therefore the system won't work. Therefore the peaks can either all be odd or contain 1 odd number, because in our starting pool there are 5 odd numbers.
From there I identified 10 cases which fall under these criteria, but only 6 of them actually worked and I am trying to find out why the others didn't.

 

[mfb]

A+B+C+D=1+2+...+9=45, so A+B+C=15 and D=30.
That gives 8 cases for A,B,C.

Spoiler

4 of them have a 5, and I think this leads to 16 results.
2 of them have a 4 but no 5, I count 6 results.
2 of them have a 6 but no 4, I think those are not possible.

This would lead to a total of 22 results. If we count rotated and mirrored solutions as separate, multiply this number by 6 to get 132 results.

 

[lendav_rott]

Oh snap, I might have expressed the assignment confusingly.
http://www.upload.ee/image/3586812/Puzzle22.jpg
this is what I mean, I've added one of my solutions when A,B,C are 9,2,4 (it works in 2 ways, since 8 and 5+3 are interchangeable.)

Is that how you understood the problem, mfb?

 

[mfb]

That's how I understood the problem, too.
(A,B,C)=(9,4,2) has the additional solutions y=7, x=81, z=356 and y=7, x=63, z=158.

And as you can see, A+B+C=15 - that is true for all solutions.

 

[lendav_rott]

Ah, in that case - the 10 solutions I found are all with a unique combination of ABC, not counting the variations right now, but 6 of the combinations work and not all of them have variations and I'm trying to figure out why that is.

Yes, I discovered that A+B+C is 15 and D is 30(I did suspect it, but I wasn't sure so I went with the even/odd theory). One of my solutions yielded 2, 6, 7 as ABC, which is 15, but the triangle can't be solved so there has to be more to the story. The 7-6 side needs a 7, which can only be made of 3, 4 - the 7,2 side needs 11, which is no longer possible.

 

[mathal]

I agree with mfb's number of solutions. Since you mentioned the A+B+C=15 the 'magic square' makes a good visual tool.

8 1 6
3 5 7
4 9 2


Each vertical,horizontal and diagonal line adds to 15 and the remaining numbers are easy to
keep track of.

mathal

 

[mathal]

Since I have naught to do right now and the weather is not great whatsoever, I'm going through a Math problem book I found covered in dust sitting on the shelf at my folks' - it's from 1960 entitled (if word for word translation) "Mathematical Sharpness" by B. Kordemsky originally in Russian, but I do my best to translate for you

I have an English translation of the book in my library.
The English title is The Moscow Puzzles by Boris A. Kordemsky.
It was published in 1972 by Charles Scribner's Sons ,translated by Albert Parry, Professor Emeritus of Russian Civilization and Langauge at Colgate College.
The book was edited by Martin Gardner, who was then the editor of Mathematical Games Department at the Scientific American Magazine.
A very enjoyable and interesting book.
mathal

 

[lendav_rott]

I agree with mfb's number of solutions. Since you mentioned the A+B+C=15 the 'magic square' makes a good visual tool.

8 1 6
3 5 7
4 9 2


Each vertical,horizontal and diagonal line adds to 15 and the remaining numbers are easy to
keep track of.

mathal

Are you saying all rows columns and diagonals make for possible combinations for ABC? If so, row 3 isn't working.

 

[mathal]

Are you saying all rows columns and diagonals make for possible combinations for ABC? If so, row 3 isn't working.

The display of the numbers this way just makes it easier to see all the possible A+B+C=15 at once.
These are the 8 cases for A,B,C that mfb referred to outside his spoiler text. Neither mfb nor I have contended that all 8 cases have solutions to the puzzle.

mathal

Spoiler

No. the two lines with a 6 and with no 5 in it do not have solutions.
6 7 2 requires a sum of 7 on one side which forces a 4 3 here and the other two sides 12 and 13 can't be made.
8 1 6 requires a sum of 6 on one side which forces 2 4 here and the other two sides 11 and 13 can't be made.

 

[mathal]

One further note on puzzle 22 in the book. In my English translation the puzzle starts with a simpler problem. The 3 points on the triangle are 1 2 and 3 and the sum to be arrived at is given as 17. By extension the opposite extreme is 7,8 and 9 on the 3 points and the sum is easily seen to be 23. [(7+8+9)*2 +1+2+3+4+5+6]/3.
The full set of embedded puzzles then goes from a sum of 17 to 23.

mathal