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6 solutions
Nice observation :)
it's nice solution
Super observation sir
I slice them into 3 5 ∣ 1 4 ∣ 9 1 . The three number in each slice has GCD at 7, but in different "hundreds" scale, so I conclude it NOT a prime. To prove that, expand the number to ( 3 5 × 1 0 4 ) + ( 1 4 × 1 0 2 ) + 9 1 Or, in an incomplete factorization, which proves the number is NOT a prime 7 × ( ( 5 × 1 0 4 ) + ( 2 × 1 0 2 ) + 1 3 )
491 - 351 = 140 is divisible by 7 hence the number itself is.
1001 = 7×11×13
351491 = 351351 + 140 = 351×1001 + 140
This is great :)
If asked a question like this (and expected to solve without a computer), the most likely solution is no. :)
Showing a number is prime is harder than showing that it is not.
351491's prime factorization is: 7 * 149 * 337.
Therefore, 351491 is not a prime number.
I feel sorry to you, you didn't get the maximum enjoyment :p
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True - not this time . Don't feel sorry for me though. :-) I get maximum enjoyment out of music and other math problems that are not easily solved by a computer.
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Haha, glad to hear it, Warren!
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@Roberto Nicolaides – Thinking about this problem a little more deeply now. could it be proved that is not prime by a brute force - division - 1,2,3,4,5,6,7 .. until you hit 7. That would take much longer then your solution -though?. Also wondering if the method that you used to discover if this is a prime is broadly applicable?
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@Warren Cowley – Wow, good question - deep thinking indeed!
The brute force method will certainly work, it would just be a bore to do by hand in this case, you're right.
As for applying it elsewhere, this is a great idea!
One thing to notice is that, unfortunately, the converse is not true. In other words, just because something is divisible by some prime p that we can use this method (assuming that we have some non-trivial splitting) to check if its prime.
For example
9 1 = 7 ∗ 1 3 but we can't split 9 1 into smaller parts that are divisible 7 .
One relatively cool thing it is useful for is for creating massive multiples of primes easily by "gluing" lots of multiples of primes next to eachother.
For example we can put get a load of multiples of 1 3 such as 1 3 , 3 9 , 1 4 3 and glue them together how ever we want, eg 1 4 3 3 9 1 3 and we know that that is prime.
It could be very interesting find out a formula for the maximum amounts of splits each number can have given a prime divisor of itself and see if this follows a cool pattern?
I expect maybe there are some really cool applications that we can use this for that I don't have the imagination to see right now (perhaps into checking if polynomials or knots are irreducible too?).
If you have any cool ideas, be sure to send them my way!
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@Roberto Nicolaides – This is maximum enjoyment! I like the idea of creating massive multiples of primes by gluing them together. Seems like a formula for this might be useful in cryptography in some way? I will be thinking about this now and I will post my discoveries. I am thinking - something can be programmed if it can be reduced to a formula and an algorithm. It might be reinventing the wheel in some way, though- as I do not have a lot of expertise in this area.
@Roberto Nicolaides – Actually maybe not cryptography but discovering large primes? I don't know- my breadth of knowledge in this area is minuscule and I am pretty sure someone in the past has already tread this ground. Warrants further inspection though.
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@Warren Cowley – "Warrants further inspection though." Agreed, sounds like you're full of good ideas and a great imagination. Be sure to keep me updated :)
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@Roberto Nicolaides – Thank you for the compliment, Roberto! I will keep you updated. Thanks for the problems on your account here at Brilliant as well.
Nice, how did you get the prime factorisation?
No , 351491 isn't a prime number , it is divisible by 7 .
If we have some number, say 1 7 3 4 1 8 7 , and we can put some 'splittings' in between some of digits like so
1 7 ¦ 3 4 ¦ 1 8 7 where each new number within some split is divisible by a prime (in this case 1 7 ) then the number itself cannot be a prime. This is because we can write
1 7 3 4 1 8 7 = ( 1 7 ) 1 0 5 + ( 3 4 ) 1 0 3 + ( 1 8 7 )
and each term in the right-hand side is divisible by 17 hence the number itself is.
This generalises nicely to all natural numbers.
Using this idea we can look at splitting the given number 3 5 1 4 9 1 into splits with a common prime factor.
One way we can do this 3 5 ¦ 1 4 ¦ 9 1 and since 7 divides all the splits it must be a divisor!
So 3 5 1 4 9 1 itself is not prime!