Brainsteem: A Prime STEEMIT Number Puzzle

in mathematics •  7 years ago  (edited)

I just made this number puzzle this afternoon to kick off my writing on Steemit. I hope you find it amusing.

Take the word STEEMIT and replace each letter with a distinct numeral from 0 to 9 to create a 7-digit number. Each repeating letter must be replaced by the same numeral and the letter S cannot be zero.

What is the smallest such number that is a prime number?

For example, if STEEMIT --> 3577245 then this is not a prime number, as divisible by 5.

Bonus upvote if you include a method with your solution. There is a logic to solve this more quickly than random!


Brainsteem Pro: I think the above is quite doable with pen, paper and a prime number checker, but for those of you who enjoy a computational challenge, try finding every possible prime number.


I have created many mathematical problems and will slowly be adding them to Steemit. They will range in difficulty from primary competitions and national challenges to pre-Olympiad questions and some computational mathematics. There will also be other kinds of puzzles, such as on logic or word puzzles, plus reviews of any clever online games; pretty much anything I find amusing. This also gives me the opportunity to freshen them up and tweak them a little.

I have started using the #brainsteem tag so that they will be easy to find - for me too! 

Please follow me @rycharde to get the latest puzzles in your feed.

Enjoy!

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1300253, is it the smallest one?

What method did you use? Just to maybe help others learn.

In order to make it smallest, I put 1 in the first digit
then the final digit can only be 3,7 or 9
So the smallest should be 13EEMI3
E should be 0 to keep it smallest
then MI I just do it by trial and error ;)

  ·  7 years ago (edited)

@kenchung I was going to use a similar method. However, you can actually cut down what M I is even more. Im all for cutting down as much work as you can.

given you got 1300MI3, you can also say that MI can not add up to 2,5,8,11,14, or 17 because the other numbers add up to 5. And we know that any number who's sum of digits is divisible by 3 is also divisible by 3. Combine that with no repeating numbers, the set to try for MI becomes much smaller.

The set of possible solutions to check becomes {24,25,27,28,42,45,46,48,49,52,54,57,58,64,67,69,72,73,75,76,78,79,82,84,85,87,94,96,97}, which is 29% of the work than going through all 100. In fact, you come up with the solution on the second try, thus could have only done 4% of the work needed :)

Sorry, had to make a small edit and the layout collapsed! Quickly re-edited with html. Apologies if you saw it look a mess.

BTW forgot to mention, I'll upvote all correct answers and any interesting comments extending the original question. It also means I too can write interesting comments rather than "Correct!" all the time.

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