New answers tagged formation-of-numbers
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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:
...
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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)
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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 ...
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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.
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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) =...
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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:
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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
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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 ...
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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
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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 ...
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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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