Integer numbers only

Consider

n^10 – n^5 = 2

What is the integer solution? (That is, no square roots, no complex numbers, just integer numbers)

You can find the solution here.

2023

Of course I wish all my readers a happy 2023!

1) 2023^**/*****
Find two numbers a and b such that a^2 – b^2 = 2023.

2) 2023^**/*****
Show that there is just one pair.

3) Difference^*/*****
How much is 2023² – 2022²?
This puzzle is an adaption of https://www.youtube.com/watch?v=NOIw0VZX1lc

Christmas was somehow more busy than I expected, so I apologize for not posting these puzzles earlier. However, here are some more 2023 brainteasers:
1) https://mathequalslove.net/2023-puzzle/^*/*****
Sarah Carter made a puzzle by writing out the year with digital digits and cutting it up.

2) https://paulmotwani.com/2023/01/02/blog-post-145-happy-new-year-brainteasers-%F0%9F%98%8A%E2%99%A5%F0%9F%98%8A/^**/*****
Paul A. Motwani brings a nice puzzle about a product and sum, which differ 2023.

New puzzles are published at least once a month on Fridays, usually the 1rst and / or third Friday of the month. 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.

Three

Use 0, 0 and 0 to make 6
Use 1, 1 and 1 to make 6
Use 2, 2 and 2 to make 6
Use 3, 3 and 3 to make 6
Use 4, 4 and 4 to make 6
Use 5, 5 and 5 to make 6
Use 6, 6 and 6 to make 6
Use 7, 7 and 7 to make 6
Use 8, 8 and 8 to make 6
Use 9, 9 and 9 to make 6
Use 10, 10 and 10 to make 6

You can use the numbers and any mathematical function such as sqrt() (√). You are not allowed to introduce other numbers.

An example of a solution could look like: 3×3-3 = 6

A new puzzle is published at least once a month on Friday. You can check your solution here.

Did you like this puzzle? The good news is I wrote a booklett with dozens of these puzzles, you van find it here.

Quento (2)

In august 2020 I published a couple of puzzles which in the Dutch daily newspaper Algemeen Dagblad (ad.nl) are called Quento.

On the left side you see some numbers and arithmetic signs. On the right you see a couple of ‘answers’. Your task is to make valid calculations by travelling through the grid and ending up with one of the answers.

Of course there is a website, and an app for both Android and Ipad. Personally I think the exercises are too easy, though no doubt they can be increased by adding size and moving to higher numbers, as shown in our 4th problem. In the app I did see higher numbers, in the sense of multiples of 3, 4, 5 , and so on, but I didn’t see larger sizes, as in our fourth problem. That may be because I didn’t purchase the app and just used the free version.

Quento 5^*/*****

Quento 6*/*****

Quento 7*/*****

You can check your solutions at here.

Some thoughts about Quento
1) The problem “find the exercise leading to the answer” sounds like it is a problem in inductive logic. But alas it is not. There are a very finite number of routes leading to a possible answer.

2) A good question is: how many routes are the through a 3×3 grid? Or more general how many routes are there through an nxm grid?

Let’s start with a 2×2 grid, consisting of four numbers and foru +/- signs in between the numbers.
If we have them arranged as
A+B
+ +
C+D,
we have A+B, A+C, B+D and C+D or 4 routes. For every minus sign we do not just have A-B, but also B-A, giving an extra 4 routes of length 2, making 4-8 routes of length 2.
In addition, there are 4 routes of length 3, and 1 route of length 4 ( taking only + signs).
With minus signs, there are an additional 4 routes of length 3, but suddenly 4 of length 4.
That gives a total of 4-8 routes of length 2, 4-8 routes of length 3, and 1-5 routes of length 4, making 17 in total.

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.