+0

I don't actually need help with this, I have done this already, but I'm curious to see what other people's answers are.

0
214
3

So, I got an assignment earlier, what I had to do was make an equation (probably the wrong wording, I'm really tired), that must contain a 1, a 2, a 3, and a 4, and only one of each number (also, exponents, square roots, and other math symbols are allowed, but are not required). That probably doesn't make sense, but I'll try to explain it.

We had to get them to result in answers between (and including) 0 to 20. Here are some examples

(1+4)-(2+3) = 0

1+2+3+4=10

4²+1³ = 17

4²+1+3 = 20

Again, I have already done this, I don't actually need help, I'm just curious as to what answers you can come up with.

Have an awesome rest of the day :-) ☺.

Edit: changed a minus to a plus, but I doubt anyone will see that.

Aug 22, 2018
edited by Guest  Aug 22, 2018
edited by Guest  Aug 27, 2018

#1
+1

$$2+3-4-1=0$$

$$\frac{2-1}{4-3}=1\\ 4-3-1+2=2\\ \frac{2+1}{4-3}=3\\ \frac{4}{2}+3-1=4\\ 4+3-1*2=5\\ \frac{4}{2}+1+3=6\\ (4+3)(2-1)=7\\ (3-1)^2+4=8\\ 1*2+3+4=9\\ 1+2+3+4=10\\ 2*3+4+1=11\\ 2^3+4*1=12\\ 2^3+4+1=13\\ 2*(3+4)*1=14\\ 3*(4+2-1)=15\\ 4*(3+2-1)=16\\ 3*(2+4)-1=17\\ 3*(2+4)*1=18\\ (2+3)*4-1=19\\ 1*(2+3)*4=20$$

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Aug 22, 2018
edited by Melody  Aug 22, 2018
#3
0

Oh, I didn't notice the reply button, oops, but thanks.

Guest Aug 22, 2018
#2
+1

Nice, I knew there were far more than just 1 possible answer. Thanks Melody (and anyone else who answers). Also, I don't know how to set it as the best answer, though I'm at school now, not home, so I'm probably not able to do that.

Aug 22, 2018