All Questions
Tagged with formation-of-numbers number-theory
29
questions
18
votes
7
answers
6k
views
Make 2 0 2 2 2 0 2 2 = 2022 [closed]
Inspired by this puzzle, I've come up with the following:
Can you find a way to make
$2 \; 0 \; 2 \; 2 \; 2 \; 0 \; 2 \; 2 = 2022$
by only adding any of the following operations or symbols:
$+,\ -,\...
user80184
10
votes
4
answers
480
views
Not so boring numbers
This is inspired by this and this and more similar ones.
Let us consider formation of a given positive integer N by the following rules
You may use only one digit, which one is your choice and you ...
- 10.3k
8
votes
4
answers
904
views
Sum of digits of sum of digits of sum of digits
The following question was asked in a competitive exam for which I am preparing and I was unable to solve it (in fact I am completely clueless about it).
So, I am asking for help here.
Given a number ...
- 687
10
votes
0
answers
512
views
I'm a number with a special product, so name me when you think you've got it
In the blocks that come before,
their special product tells us more.
To guess my scheme you'll need calculation,
but only little tests of recreation.
Each block contains atoms strong as Thor,
...
- 2,296
11
votes
4
answers
931
views
Making the whole set into primes
Let's say you start with a set of sequential integers starting from 2, so: $ 2, 3, 4, 5, \dots, N $ for some $ N > 2. $
The goal is to use identical basic arithmetic operations ($ +, -, \times, \...
- 113
11
votes
4
answers
5k
views
Create the numbers 1 - 30 using the digits 2, 0, 1, 9 in this particular order!
Inspired by the last year's "2018 four 4s challenge", I thought it's time to welcome 2019 by a similar challenge. This time you have to use the digits 2, 0, 1, 9 in this particular order to create the ...
- 1,370
9
votes
2
answers
2k
views
Ten-digit number that satisfy divisibilty rules for 2,3,4,5,6,7,8,9,10&11
Question:
Arrange the digits 1,2,3,4,5,6,7,8,9,0 to make a ten-digit Number that satisfies all of the divisibility rules for 2,3,4,5,6,8,9,10,&11. BONUS: make the number also divisible by 7
- 101
5
votes
8
answers
621
views
Make $1,\dots,15$ using $3, 9, 9, 9$
This was inspired by many puzzles that use three/four numbers to create other numbers. I chose these numbers in particular because of this post.
Can you find a way to make all the natural numbers ...
- 6,682
69
votes
24
answers
23k
views
Make 0 0 0 0 = 8
Can you find a way to make:
$0\ 0 \ 0 \ 0 = 8$
by adding any operations or symbols? You can use only these symbols:
$+,\ -,\ *,\ !,\ /,\ \hat\, ,\ ()$.
It is limited to this list, and ...
- 3,181
7
votes
3
answers
325
views
Is this number unique?
Inspired by Interview Question or Pathbreaking puzzle and A121808.
Start with $1$, and count the number of times $1$ occurs, and report this in the format 'number of ones:1', i.e. the next term is $...
- 35.6k
10
votes
6
answers
2k
views
Maximize the number of factorials in your solution to 6 5 4 3 = 1
Using the same rules as Make 6 5 4 3 = 1, but maximize the number of factorials. You may not take the factorial of 1 or 2.
This is harder than it looks. A great answer has six factorials, an ...
- 279
11
votes
11
answers
10k
views
Make 6 5 4 3 = 81
Can you find a way to make:
6 5 4 3 = 81
by concatenation and/or adding any of (and only) these mathematical operators:
+
-
×
!
÷
^
standard parentheses ()
You cannot add other numbers to the ...
- 21.2k
3
votes
9
answers
1k
views
Make 6 5 4 3 = 1
Can you find a way to make:
6 5 4 3 = 1
by concatenation and/or adding any of (and only) these mathematical operators:
+
-
×
!
÷
^
standard parentheses ()
You cannot add other numbers to the ...
- 21.2k
20
votes
10
answers
14k
views
Make 5 5 5 5 = 19 [closed]
Can you find a way to make:
$5\ 5 \ 5 \ 5 = 19$
by adding any operations or symbols? You can use only these symbols:
$+,\ -,\ *,\ !,\ /,\ \hat\, ,\ ()$.
It is limited to this list, and ...
- 3,181
0
votes
2
answers
245
views
Factor the number 23 into four numbers $a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}$ [closed]
We know that $23$ is a prime number nonetheless, I'm asking to find 4 numbers $a,b,c,d > 0$ such that $23$ factors.
$$ 23 = A \times B \times C \times D \text{ with } A,B,C,D = a + b \sqrt{2} + c \...
- 269