Your task is to create every number from 1 to 50. You can use only the digits 1, 2, 3 and 4 once in each and the operations + – * / . So, here goes… 1 to 50, using only 1, 2, 3 and 4 and the basic maths operators.
How many combinations can you make with the numbers 1, 2, 3?
Starting with 1 2 3 we can form combinations of size 1 2 or 3. n! r!(n − r)! That is a total of 7 combinations. (If we wish to count choosing 0 items — the empty combination — as a combination, then we must add 1 way of doing that.)
When do you use the digits 1, 2, 3, 4?
THE RULES: You MUST use the digits 1, 2, 3, and 4 in each equation. You MAY NOT use any other digits (0, 5, 6, etc.) You must use each digit (1-4) once and only once in the equation. For example, you MAY use 1+2+3+4, but NOT 1+1+2+3+4, and NOT 1+2+3 You may NOT create 2 digit numbers.
Is the number 3 the same as 1 2 3?
In a permutation order matters, so the permutation (1 2 3) is not the same as (2 1 3). In a combination order does not matter so the combination (2 3) is the same as (3 2). The question does not say whether we are allowed repetition or not. So we do not know whether we are to count (1 1 3) as a combination of 3.
Why are 1, 2, 3 and 4 considered small numbers?
As the logical extension, I attempted to carry on past 50. It becomes increasingly difficult, since 1, 2, 3 and 4 are all small numbers, and the combinations of those small digits become less useful in making specific larger values (especially the prime numbers).
Can you use only 1, 2, 3 and 4 in maths?
You can use only the digits 1, 2, 3 and 4 once in each and the operations + – * / . So, here goes… 1 to 50, using only 1, 2, 3 and 4 and the basic maths operators. Some of the answers seem a little repetitive or derivative (look at 38 through 43), and in some cases I found alternative answers afterwards.
When do you put placeholder 0 in binary calculator?
Since the only values used are 0 and 1, the results that must be added are either the same as the first term, or 0. Note that in each subsequent row, placeholder 0’s need to be added, and the value shifted to the left, just like in decimal multiplication.
Can you create all numbers from 0 to 100?
Create all numbers from 0-100 only using 1,2,3,4 and 5. No repeats and you have to use each number. Also, you can use any operation. I’ve only gotten to 50 by pure brute force. I think that this might be fun for you puzzlers!
Can you make a puzzle with only 50 numbers?
The puzzle is from an old maths textbook pitched at 12-14 year olds, but has potential for expansion into more difficult territory. Your task is to create every number from 1 to 50. You can use only the numbers 1, 2, 3 and 4 once in each and the operations + – x / .
Is there a way to make 10 from the numbers?
Make 10 from the numbers 1, 1, 5, 8. You can use the operations + – × ÷ (). You have to use all the numbers, and use each number exactly once. The operations can be repeated (like 1 + 1 + 5 + 8) and you don’t have to use each operation. However, you must use only these operations. You cannot use exponents so 8 + 1 + 1 5 = 10 is not a solution.
How to make numbers using just four digits?
Make as many numbers as you can using just four digits! There is a simple arithmetic puzzle, where you are given four digits, and you are told to build as many numbers as possible using just those four digits and the basic operations.
Is it possible to find solutions for 1, 2, 3 and 4?
It becomes increasingly difficult, since 1, 2, 3 and 4 are all small numbers, and the combinations of those small digits become less useful in making specific larger values (especially the prime numbers). However, if we expand the rules to allow ! (factorial) and decimal points, then this enables us to find solutions for 57 (for example).
