Catch those naughty little primes
3 4 5 ! + 2 , 3 4 5 ! + 3 , 3 4 5 ! + 4 , … , 3 4 5 ! + 3 4 5
Find the number of primes among the numbers above.
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7 solutions
I didn't understand the question. I read the problem as containing only one number.
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I agree. The commas spacing don't make it clear to see that it's a list of numbers demarcated by the commas. Was confused for a bit as I thought the comma was part of the number
Even if the comma is part of a number, the problem has just a single number which is the sum of some numbers. The sum is not a prime number as it is clearly a product like 345 * (.......).
So, the answer is none of them is prime.
I also think the problem was very poorly written.
But 2and 3 are prime
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Yes, but 345! is divisible by every number less than or equal to it, including 2 and 3.
In the future, with a problem like this, add a comment to the problem specifying that the commas do not delineate a range of numbers. I read I tried to read the problem as a sequence of numbers that started with a large number , continued with an undetermined number of numbers, and ended with the smaller number , and came up with an answer that could not be determined . So I just guessed no prime numbers and then read the documentation to understand the problem. The problem is poorly written. Remember that some of your visitors do not understand every bit of nomenclature.
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Saw this problem again in my feed, and still found it confusing.
I initially saw it as
345! plus 2,345! plus 3,345! plus 4, which is ONE ITEM
The Brilliant Editors should change those commas to semicolons for better readability.
I could not understand the question....
Im not sure if I'm right but aren't they all non primes because ! Means to times 1 by 2 by 3.....by 345 so it's not prime because it's divisible by what ever number was factorizing it? I'm not sure though
Yes factorials. 345!=1×2×3×4×...×343×344×345. But also notice before each comma it says like 345!+3, 345!+4 etc
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In one of the previous questions it explained that if within a series of numbers multiplied are "5s" and "2s" then the final product ends in "0", therefore the digits added ( _ !+n) to each 345! are what would have made a difference, right?
345! + x (with 2<=x<=345)
= 1.2. ... x ... .345 + x
= (1.2. ... (x-1).(x+1)... .345 + 1) . x ==> which is a multiplication ... so factors ... so no prime
f.i. 345! + 2 = 1.2.3...345 + 2 = (1.3.4...345 + 1).2
f.i. 345! + 3 = 1.2.3...345 + 3 = (1.2.4...345 + 1).3
345! + 344 = 1.2.3...343.344.345 + 344 = (1.2.3...343.345+1).344
345! + 345 = 1.2.3...343.344.345 + 345 = (1.2.3...343.344+1).345
Interestingly, this allows you to construct any length of string of consecutive non-prime numbers. If you started with 1 000 001! + 2 and went up to 1 000 001! + 1 000 001 you would have a string of 1 000 000. This could go as mind-bogglingly long as you wanted it. ( We are talking about inconceivably big numbers here ).
This is how I solved it without the need of a paper , just mind. To be sure: take 345x2 = not a prime (definitely). If we take 345 +X where x is a natural number and can be between 0 and 346 example: 345 then it goes to be 344! 345+345= (344+1) 345=not even. But for whatever number is X to be (a N number) it will interact with one of the numbers of the 345! like shown up how. Now it's clear.
By definition : n! = 2 * 3 * ... * (n-1) * n
[ 2 ; n ] are divisors of n! so k n ! , is always an integer when 2 ≤ k ≤ n.
for 2 ≤ k ≤ n,
n ! = k ∗ k n !
n ! + k = k ∗ k n ! + k
n ! + k = k ( k n ! + 1 )
n ! + k is necessarily a multiple of k.
Therefore there is never a prime between n! + 2 and n! + n
Similar to @Kazem Sepehrinia
Because 2 ≤ n ≤ 3 4 5 is a divisor of 3 4 5 ! so 3 4 5 ! + n = n ( n 3 4 5 ! + 1 )
Which is a composite so the answer is None of them are prime
Note
- A composite is a product a b with a , b ≥ 2 , a , b ∈ N
Coz every term has common factor , the number that is being added is the common factor... So atleast it will be divisible by that number
Hence no prime number are there!
For 2 ≤ n ≤ 3 4 5 , n is a divisor of 3 4 5 ! and 3 4 5 ! + n = n ( n 3 4 5 ! + 1 ) Which is a product of two integers greater than 1 .