- Thread starter shahar
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I think what you are asking is this common puzzle: Using each of the digits 1 through 9 exactly once, fill in the blanks to make a correct equation:

_ _ _ + _ _ _ = _ _ _

That is, in correct English,

Using each of the digits 1 through 9 exactly once, create three number**s** such that one number is the sum of the **other** two numbers.

I think it quickly becomes clear that the numbers must have 3 digits each.

Now, what have you tried? This is a puzzle commonly given to students who are just learning to add; you should be able to do better with more experience, but it is still mostly trial and error.

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As you wrote: The digits 1 to 9...

Thanks..

If I try elaborate to more number (with digit 1-9) that are also have thing rule (one number have the sum of others that every number have exactly digits different from the the others so that

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Are you asking if you could change the problem to involve numbers with more digits, but still use each digit only once? That would be stupidity, not a paradox -- so I assume that is not what you mean.

Maybe if you gave an example, it would help; but really, you need to get help with your English in order to be understood. (I know your English is far better than my Hebrew, but that isn't saying much.)

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i show my way to check;

[Why it is wrong?]

First of all, I try to explain myself.

1 - I have a 3 number.

2. One of the sum of the 2 other numbers.

3. Each humber have a unique sequence of digit order that not have no repeat in the sequence.

Term to checks:

4. How many numbers have three digits? Becuase there are only 3 digits to the maximum is 3! but the posibile number is 9 digits, So there only (9 - 3!).

*There secoond question if I increase the number of digits in each number but it repeat on it sellf more than one time but in the optimize way that possible - Is it a pradox?

A. Can term number 4 help find the number?

B. The question that mark with *.

I think the above is wrong.4. How many numbers have three digits? Becuase there are only 3 digits to the maximum is 3! but the posibile number is 9 digits, So there only (9 - 3!).

The number of ways to arrange 3 distinct digits is 3!

The number of ways to choose 3 digits from {1,2,..,8,9} is \(\displaystyle \binom{9}{3}\)

The number of 3 digit combinations possible using {1,2..,8,9} is \(\displaystyle \, 3! \times \binom{9}{3}=3! \times \frac{9!}{3! (9-3)!}=504\)

I don't understand the rest of your post. Can you give examples? Are you searching for all combinations like this:-

124+659=783

157+329=486

193+275=468

...

I think you are asking why there isn't a systematic way (an algorithm) to find all combinations more quickly than trying all the combinations (brute force).I still not understand, Why isn't there a way other thantrial and errormethod to find the numbers?

Why isn't be a rule or a systematic way to claculate it, than guess (or use program computer :~|) to find the numbers?

To make the algorithm more efficient usually involves making it more complicated. This observation could be used to speed up the search...

Code:

```
If this pattern fits:-
A B C
+ D E F
=====
G H I
then these patterns also fit:-
A B F A E C D B C
+ D E C + D B F + A E F
===== ===== =====
G H I G H I G H I
```

BUT not all problems can be solved quickly. Modern encryption relies on this fact. I recommend that you look up "NP-complete"., "NP hard", etc.

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When I said "trial and error", I meantI still not understand, Why isn't there a way other thantrial and errormethod to find the numbers?

Why isn't there a rule or a systematic way to calculate it, other than guess (or use program computer :~|) to find the numbers?

The thing that makes it hard is the requirement that no digits repeat, which means that algebra, while perhaps useful to cut down the number of attempts, can't go all the way to an answer.

One thing you can use to restrict possibilities to test is divisibility rules. For example, the sum of the digits in the entire sum is 45, which is a multiple of 9. If we look at the equation (mod 9), that is, think only about remainders on division by 9, we can see that we are dividing the nine digits into two sets whose sums differ by a multiple of 9, which implies (if you think about it long enough) that the sum is itself a multiple of 9. That gives you fewer numbers to try, but you still have to check many of them.

And I doubt there is a direct way to count the number of solutions; it turns out to be 168, according to this page.