1-2-3-4 POW
Problem Statement:
For this problems, we had to find 1-2-3-4 equations for the numbers 1-25. A 1-2-3-4 equation is an expression that uses those digits only once. To find these equations we could use any arithmetic operation: exponents, radicals, factorials, parentheses/ brackets, and we could also juxtapose the numbers (put them together as in: 2 and 3 become 23).
Process Description:
What I did to solve this problem, was to see all the combinations I could come up with and see how I could make each number. Constantly using the tool of conjecture and test I went through many, many renditions of problems trying to get the number I wanted. As I finished the majority of them, I realized that the only operations that I was using was the simple ones (add, subtract, multiply, and divide). Since I already got that far, with little struggles, I decided to finish the rest of them with those simple operations. I felt like I was underachieving by "going the easy way out," so for the extension problems I did below, I decided to try to challenge myself and use some of the more difficult operations.
Solutions:
1 = (4 - 3) x (2 - 1)
2 = 4 - 3 + 2 - 1
3 = (4 - 1) x (3 - 2)
4 = 4 + 3 - 2 - 1
5 = (4 + 1) x (3 - 2)
6 = 4 + 3 - 2 + 1
7 = (4 + 3) x (2 - 1)
8 = (4 - 2) x (3 + 1)
9 = 3 x 2 + 4 - 1
10 = 4 + 3 + 2 + 1
11 = (4 x 2 + 3) x 1
12 = 4 x 2 + 3 + 1
13 = (4 + 1) x 2 + 3
14 = (4 x 3 + 2) x 1
15 = 4 x 3 + 2 + 1
16 = (4 + 3 + 1) x 2
17 = (4 + 1) x 3 + 2
18 = (4 + 2) x 3 x 1
19 = 4 x (3 + 2) - 1
20 = 4 x (3 + 2) x 1
21 = 4 x (3 + 2) + 1
22 = (4 x 3 - 1) x 2
23 = 4 x 3 x 2 - 1
24 = 4 x 3 x 2 x 1
25 = 4 x 3 x 2 + 1
Extensions:
One extension to the problem I came up with, is that you can only repeat digits if they are exponents. This lets you use bigger numbers and allows for longer equations. Some of the answers I found for this are:
1: (3^2)-4x2x1
2: 21-(4^2)-3
3: (4^2)-(2^2)x(3^2)x1
4: 1+ [(2^2)-4]+3
5: (4^2)-(3^2)-2x1...
Self-Assessment and Reflection:
During this problem, I learned that you can use a small amount of digits to form equations for many numbers if not all numbers. The ways you set up the equations and the mathematical tools you use in the equations, will impact the final result heavily. The problem also made me think more thoroughly and have to constantly use the habit of a mathematician: 'conjecture and test.' I was consistently setting up multiple equations trying to get the number I needed to find. Another habit of a mathematician I think I have conquered through this problem was 'stay organized.' To complete this problem, I had to organize all of my equations to see what I did to get that specific answer. I'm glad I did, because on a few of the equations, I used part of the previous equation to cooperate it with the next problem. For example, on number 19, I came up with the equation 4 x (3 + 2) - 1. Instead of trying to come up with a completely different equation, I just substituted the 'minus 1,' with a 'times 1,' so it equals 20 instead. Coincidentally, this method also works for number 21. I just substituted the 'times 1,' with 'plus 1,' so it equals 21. I think out of 10, I would give myself between a 9 and a 10 in this problem, because I honestly think I was persistent with my method and I used the tools that were necessary.
For this problems, we had to find 1-2-3-4 equations for the numbers 1-25. A 1-2-3-4 equation is an expression that uses those digits only once. To find these equations we could use any arithmetic operation: exponents, radicals, factorials, parentheses/ brackets, and we could also juxtapose the numbers (put them together as in: 2 and 3 become 23).
Process Description:
What I did to solve this problem, was to see all the combinations I could come up with and see how I could make each number. Constantly using the tool of conjecture and test I went through many, many renditions of problems trying to get the number I wanted. As I finished the majority of them, I realized that the only operations that I was using was the simple ones (add, subtract, multiply, and divide). Since I already got that far, with little struggles, I decided to finish the rest of them with those simple operations. I felt like I was underachieving by "going the easy way out," so for the extension problems I did below, I decided to try to challenge myself and use some of the more difficult operations.
Solutions:
1 = (4 - 3) x (2 - 1)
2 = 4 - 3 + 2 - 1
3 = (4 - 1) x (3 - 2)
4 = 4 + 3 - 2 - 1
5 = (4 + 1) x (3 - 2)
6 = 4 + 3 - 2 + 1
7 = (4 + 3) x (2 - 1)
8 = (4 - 2) x (3 + 1)
9 = 3 x 2 + 4 - 1
10 = 4 + 3 + 2 + 1
11 = (4 x 2 + 3) x 1
12 = 4 x 2 + 3 + 1
13 = (4 + 1) x 2 + 3
14 = (4 x 3 + 2) x 1
15 = 4 x 3 + 2 + 1
16 = (4 + 3 + 1) x 2
17 = (4 + 1) x 3 + 2
18 = (4 + 2) x 3 x 1
19 = 4 x (3 + 2) - 1
20 = 4 x (3 + 2) x 1
21 = 4 x (3 + 2) + 1
22 = (4 x 3 - 1) x 2
23 = 4 x 3 x 2 - 1
24 = 4 x 3 x 2 x 1
25 = 4 x 3 x 2 + 1
Extensions:
One extension to the problem I came up with, is that you can only repeat digits if they are exponents. This lets you use bigger numbers and allows for longer equations. Some of the answers I found for this are:
1: (3^2)-4x2x1
2: 21-(4^2)-3
3: (4^2)-(2^2)x(3^2)x1
4: 1+ [(2^2)-4]+3
5: (4^2)-(3^2)-2x1...
Self-Assessment and Reflection:
During this problem, I learned that you can use a small amount of digits to form equations for many numbers if not all numbers. The ways you set up the equations and the mathematical tools you use in the equations, will impact the final result heavily. The problem also made me think more thoroughly and have to constantly use the habit of a mathematician: 'conjecture and test.' I was consistently setting up multiple equations trying to get the number I needed to find. Another habit of a mathematician I think I have conquered through this problem was 'stay organized.' To complete this problem, I had to organize all of my equations to see what I did to get that specific answer. I'm glad I did, because on a few of the equations, I used part of the previous equation to cooperate it with the next problem. For example, on number 19, I came up with the equation 4 x (3 + 2) - 1. Instead of trying to come up with a completely different equation, I just substituted the 'minus 1,' with a 'times 1,' so it equals 20 instead. Coincidentally, this method also works for number 21. I just substituted the 'times 1,' with 'plus 1,' so it equals 21. I think out of 10, I would give myself between a 9 and a 10 in this problem, because I honestly think I was persistent with my method and I used the tools that were necessary.