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How many ways one can put prime numbers to form 3 digit NIP?

  • Thread starter Mrencko
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Homework Statement


as listed above the question is how many and which three digit NIP can be formed whit the use of prime numbers[/B]


Homework Equations


nothing currently trying to understand[/B]


The Attempt at a Solution


well i have found at least 168 primer numbers below 1000 i mean in the range of three digit,
and grouped in three groups:
numbers of 1 digit "4"
numbers of two digit "21"
numbers of three digit ""143"
as far i know this is a permutation because order matters so 717 is diferent of 177 and 771 so
i am thinking of like a billion of ways to put those numbers to form a NIP, my question is this is even doable?
how can i start to mix this to make to the final count of how many ways one can put all those numbers to form the NIPS
***** update: i think for the three digit numbers there is a rule of 3! on each one so making 6 ways to put that number so if i multiply that for 143 this gives me 858 ways in total but i dont know if this is correct, and its just for the three digit numbers
**** second update:
i permuted every 1 digit number whit every 2 digit number
11 and 2,3,5,7 ok then 112, 211,121. so 3!=6 then 6*4 the 4 represent the 1 digit numbers
24 is the total acoding to this so 24*21 21 represents the total 2 digit numbers, this gives to me
504 but previously i ve calculated the permutation of 3 digit numbers so using the prefix "and"
504*858=432432
i dont know if i am right can you help me?
 
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Answers and Replies

  • #2
WWGD
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What is an NIP?
 
  • #3
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i guees the number for the bank and things like that only uses numbers and not letters
 
  • #4
WWGD
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Do you mean PIN, personal identification number?
 
  • #5
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yes but in my homework says NIP
 
  • #6
WWGD
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Can you find out how it is defined?
 
  • #7
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is in spanish "numero identificacion personal" check my lastest update
 
  • #8
WWGD
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Ok, did not expect Spanish with that username. Seems you have the option for 3, of 4P3= ##\frac {4!}{1!}=4!=24 ## with just one digit. Then you can have combinations of a 1-digit prime in the 1st, 2nd or 3rd spot and a two-digit prime in the remaining two spots ( if you allow this; maybe you just allow a 1-digit prime in spot 1 and a 2-digit prime afterwards or a 1-digit prime in spot 3 and a two-digit prime in the first two spots, and then consider all the 3-digit primes.
 
  • #9
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so my analisis in the update its all right? or have some flaws
 
  • #10
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my new doubt is if do i need to multiply the results?
i mean the 24 forms of the 1 digit numbers, the 504 form for the two digit numbers and 1 digit number and the 858 form of the three digit numbers?
24*504*858=?
 
  • #11
haruspex
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can be formed whit the use of prime numbers
This is too vague.
It could mean just using prime digits, or concatenating 1-, 2- and 3-digit primes.
I'm pretty sure it does not mean more convoluted uses like this:
i think for the three digit numbers there is a rule of 3! on each one so making 6 ways to put that number
If you allow that sort of thing then you can almost surely make every 3-digit number not ending in zero. Your multiply by 6 rule will in itself count duplicates, e.g.133 would be counted twice.
 
  • #12
WWGD
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This is too vague.
It could mean just using prime digits, or concatenating 1-, 2- and 3-digit primes.
I'm pretty sure it does not mean more convoluted uses like this:

If you allow that sort of thing then you can almost surely make every 3-digit number not ending in zero. Your multiply by 6 rule will in itself count duplicates, e.g.133 would be counted twice.
Yes, this is where I sort of got stuck. It seems to need a version of multinomial coefficients, you know, the Mississippi thing..
 
  • #13
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my new doubt is if do i need to multiply the results?
i mean the 24 forms of the 1 digit numbers, the 504 form for the two digit numbers and 1 digit number and the 858 form of the three digit numbers?
24*504*858=?
Look up multinomial coefficients. These help you answer, e.g., the number of permutations of a word like Mississippi ( with many repeats ) as ## \frac {11!}{2!4!4!} ##; 4 repeats for s, for for i and 4 for p.
 
  • #14
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yes i also think this is too vague, i will keep working on this and keep you updated.
 
  • #15
haruspex
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yes i also think this is too vague, i will keep working on this and keep you updated.
My best guess is that the question just means using prime digits.
 
  • #16
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1 digit prime numbers?
 
  • #17
haruspex
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1 digit prime numbers?
Yes. If it does not mean that, my next guess is using 3-digit prime numbers and concatenating 1-, and 2-digit prime numbers in either order, but certainly nothing more complicated than that.
 
  • #18
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Yes, I didn't tough about the repeating numbers in the 3 digit, ultimately I will ask my professor tomorrow and update, I only hope this don't come in the test
 
  • #19
haruspex
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Yes, I didn't tough about the repeating numbers in the 3 digit, ultimately I will ask my professor tomorrow and update, I only hope this don't come in the test
It doesn't matter about repeating digits in a three digit prime as long as you don't allow reordering its digits.
What makes my second guess complicated (and therefore unlikely) is that you have to avoid three digit and two digit primes that can be made from one digit primes.
E.g. you can make 223 as 2,2,3 or 2,23 or 223, but must only count it once.
I recommend you to take the simplest view, just using single digit primes, and answer that.
 

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