# Tag Info

### How to solve 1 2 3 4 5 = 5 4 3 2 1 (insert five pluses to make it equal)? A thorough solution needed

Here is a way to solve this problem using python: ...

### How to solve 1 2 3 4 5 = 5 4 3 2 1 (insert five pluses to make it equal)? A thorough solution needed

Actually is quite simple it’s 0+1+3+5 <> 0-+4/5-+422 +5+3+2+3+4-5 %1.5*{3+64-1+2)(c17h35COONa)
1 vote

### How to solve 1 2 3 4 5 = 5 4 3 2 1 (insert five pluses to make it equal)? A thorough solution needed

In my method I incorporate a parity check into the calculations. First off, we know that there cannot be four + on either side because that makes thone sum equal to 15 and the other having at least ...

### How to solve 1 2 3 4 5 = 5 4 3 2 1 (insert five pluses to make it equal)? A thorough solution needed

Admittedly, I finish with trial and error, but after narrowing it down to 7 calculations.
1 vote

### How to solve 1 2 3 4 5 = 5 4 3 2 1 (insert five pluses to make it equal)? A thorough solution needed

This interested me so much I created an account just to respond! Take the generalised case: x x+1 x+2 x+3 x+4 = x+4 x+3 x+2 x+1 x Try a solution at random x * (x+1) + (x+2) + (x+3) * (x+4) =...

### How to solve 1 2 3 4 5 = 5 4 3 2 1 (insert five pluses to make it equal)? A thorough solution needed

The plus sign is also a unary operator. Therefore a solution with four plus signs is also valid for five plus signs:

### How to solve 1 2 3 4 5 = 5 4 3 2 1 (insert five pluses to make it equal)? A thorough solution needed

Another way to think about it is to start with plus signs in all eight positions. The sums are equal at $15$. When you remove a plus sign you add

### How to solve 1 2 3 4 5 = 5 4 3 2 1 (insert five pluses to make it equal)? A thorough solution needed

It seemed like there should be more addition signs on the right than left because the numbers you could make on the right side by omitting plus signs are bigger on account of the bigger numbers coming ...

### How to solve 1 2 3 4 5 = 5 4 3 2 1 (insert five pluses to make it equal)? A thorough solution needed

Well, here is how I would (and did) solve it: First step: Without the « five + » constraint Second step: Getting rid of some « + » A solution we found: Extra note: finding all solutions
1 vote

### Approximate pi out of small numbers

$n=10$: Attempt 1.1: Expansion of arctan(1) [3 terms] Yeah, this might not be a great idea. We know that So we can approximate this with three terms to get Attempt 1.2: Expansion of arctan(1) [4 ...
1 vote

### 1984 - take the digits 1,9, 8 and 4 and make 246

Assuming repeating decimals are allowed: I acknowledge that the answer by PotatoLatte is nicer.

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