A King, 1000 Bottles of Wine, 10 Prisoners and a Drop of Poison

You probably know the kind of puzzle where a trader has a number of gold coins, one of which is fake and has a different weight. His only tool is a balance with two scales, and the challenge to find out which of the x coins is fake in the smallest possible number of weighings.

I am not sure who first published this puzzle, and it has spread all over the web.

Sultan Ibrahim the 512th received all his 100 gouverneurs for the wedding of his favourite son. Every one of them brought a bottle of wine for the celebrations. His wife however found an anonymous note that one of the bottles had been poisoned. Alas her information did not tel her which gouverneur or which bottle. It did state that the poison was untraceable, and that the victim would die after exactly 24 hours.
The Sultan had 10 prisoners How did he use the 10 prisoners to start the celebrations of his sons wedding the next day?

You can find the solution here..

2, 3, 4, 5

Two weeks ago we made a list of all the numbers we could make by combining the integers 1, 2, 3, and 4 with the common arithmetic operations +, -, * and /.

2, 3, 4, 5^***/*****
This week I challenge you to make the numbers 0 – 20 by using the digits 2, 3, 4, and 5 with the common arithmetic operators mentioned above.
If you are really stuck, you may use the “!” operator, but don’t use it if you can do without it.
In case you are not familiar with the !-operator:
2! = 2*1 = 2
3! = 3*2! = 3*2 = 6
4! = 4*3! = 4*6 = 24
5! =5*4! = 5*24 = 120.

New puzzles are published at least twice a month on Fridays. You can check your solution here.

1, 2, 3, 4

1) Use the digits 1,2, 3 and 4 each once to make all numbers 0 to 33. You may combine them in any way you want with +, -, * and /.
2) Proceed to 42 by also using exponents.

New puzzles are published at least twice a month on Fridays. You can check your solution here.

Alice and the pies

Alice and the March Hare had a Christmas lunch. Alice had baked 5 pies, the March Hare 3.
“It’s tea time,” the March Hare said. “So let’s eat the pies.”. The Mad Hatter popped in.
“It’s lunch time, not tea time,” Alice said. “But we can eat the pies.”
Each pie was cut into 3 parts, with one part eaten by each of the three.
“It was the best butter, you know” the March Hare said. “And there’s nothing better than the best butter, you can’t deny that.”
Alice looked surprised at him, as she didn’t understand why he made that remark.
“Well, I think they all are delicious”

At the end, the Mad Hatter thanked them, paid 8 pounds and left.

“Now that’s 5 pound for you and 3 for me,” the March Hare said.
But Alice doubted this was fair. Was Alice right?

I found this problem at https://plus.maths.org/content/sharing-cakes. Reportedly, a version of it was written by Ali ibn Abi Talib in the seventh century AD. Another version appears in Fibonacci’s famous Liber Abaci.

New puzzles are published at least twice a month on Fridays. You can check your solution here.

The numbers 1 – 9 and combinatorics.

As children, we all learned to count. Thus we rarely think about counting as ‘difficult’. Yet mathematicians have developed a special branch of mathematics for the art of counting. The branch is called Combinatorics. Typical questions in Combinatorics are:
1) In how many ways can a stack of 52 playing cards be arranged?
2) When we have a vase with 5 black and 5 white balls, in how many sequences can we pull them out?

In today’s problems, we work with the cards 1 to 9:

1) How many ways?^*/*****
It is easy to arrange these cards into 3 groups, all with the same sum:

One reason it is easy, is because there are several solutions. Withe sum of the 9 cards being 45, each of the three groups will have to have sum 15. But in how many ways exactly can we divide the cards 1-9 into three groups, all with the same sum?

2) Be creative^****/*****
Now be creative in the arrangement of your cards. In how many ways can you create 3 groups in such a way that the three groups still all have the same sum, but the sum is not 15?
Yeah, you may cheat in this problem. But your cheating is limited to arranging the cards.

3) Combinatorics (unsolved)^*****/*****
The sum of the first n cards is n(n+1)/2. To divide these number into three groups with the same sum, either n or n+1 mus be a multiple of 3. So this is not possible for n=4, 7, 10 and so on.
Here is a short list
n
2: 0
3: 0
5: 1 (5, 1-4, 2-3)
6: 1 (1-6, 2-5, 3-4)
8: 4
9: see answer to problem 1.
Now can you find a general formula for the number of possible groups?
Or for a simpler start: in how many ways can we draw cards from a series 1-n in such a way that that the sum is some given number?
Or: can you construct an algorithm that shows that the cards 1-3n (n>=2) can always be divided into three groups with the same sum?
I don’t have the answers to these questions, they just look interesting to me.

A new puzzle is published on Fridays, at least twice a month. You may check your solutions here.

Complete this multiplication

1) Complete this multiplication^**/*****
Fill in the missing numbers such that
54 x 2__ = ____8
is a correct multiplication and all the digits 0 – 9 are used exactly once.

This problem was published in the Dutch mathematics magazine Pythagoras, issue no 4 in year 9 (1970).

New puzzles are published at least twice a month on Friday.
You can find the solution here.

Fibonacci series

Today’s puzzles were inspired by Michael Jacobs “Introductory Thinking Tasks”.

You all know the Fibonacci series: a term is the sum of the two previous terms, with the first two given. The top row shows a classic sequence:

Your task is of course to fill in the empty spaces of the other two rows.

New puzzles are published at least once a month on Fridays. Solutions are published after one or more weeks. You are welcome to remark on the difficulty level of the puzzles, discuss alternate solutions, and so on. Puzzles are rated on a scale of 1 to 5 stars. You can check your solution here.

The sum of the digits

If you think that mathematics and arithmetic is something for boys, you probably never read about Shakuntala Devi. Born in 1929, as a young child she was taken by her father on road shows to display her abilities for mental calculation. At the age of six she demonstrated her abilities at the University of Mysore.

In 1977, when she must have been 48, at Southern Methodist University, she gave the 23rd root of a 201-digit number in 50 seconds.[6][4] Her answer, which was 546,372,891, was confirmed by calculations done at the US Bureau of Standards by the UNIVAC 1101 computer, for which a special program had to be written to perform such a large calculation, which took a longer time than for her to do the same.

She wrote several books, on puzzles, astrology, memory and homosexuality. What interests us here is the book “Puzzles to puzzle you”.
Here is one of them:
Which number is exactly 3 times the sum of its digits?

In her honor :
Which number is exactly 11 times the sum of its digits?

You can check your solution here

Four equal sized areas

This week we have some calculation problems for you.

1) four areas 4×6^**/*****
Divide this grid into four equal sized areas, each with the same sum.

2) four areas 4×7^**/*****
Divide this grid into four equal sized areas, each with the same sum.

You can check your solutions here.

New puzzles are posted twice a month on Friday. You are welcome to remark on the difficulty level of the puzzles, discuss alternate solutions, and so on. Puzzles are rated on a scale of 1 to 5 stars. You can check your solutions here.

What’s next?

1) Next^*/*****
What is the next number in the series 17, 72, 28, 83, 39, 94, 50, 05, 61, …

New puzzles are published at least once a month on Fridays. Solutions are published after one or more weeks. You are welcome to remark on the difficulty level of the puzzles, discuss alternate solutions, and so on. Puzzles are rated on a scale of 1 to 5 stars. You can check your solutions here.