n * (n+1)I tried a lot of numbers that’s divisible by 11 and substitute it to the n. The statement is true therefore it is not counterexample. I need a whole number if you substitute it to n, the result would be not a prime number. Please help me
If you select n = 10 or n = 21 or n = 32 what do get? Can you generalize this process (n = ?)Maybe you misunderstood my question. The formula is n (n+1) + 11 and I need a whole number that will be substitute to ”n” and the result should not be a prime number. Do you think there is a whole number that will be substitue to “n” and the result is not a prime number?
Wait wait, finally a problem that I can answer! How about 10 and 11 or even 11 and 12!!You don't need counterexampleS. It takes only one. What sort fo search have you conducted?
n * (n+1) Just looks SO tempting. Can you think of two consecutive integers where one of them happens to be divisible by 11?
A "brute strength method" is to simply start trying whole numbers:If n is a whole number, then n^2+ n + 11 is a prime number. Give me counterexamples of it
It will happen again with 21 and 22, and so on.....A "brute strength method" is to simply start trying whole numbers:
0 0+ 0+ 11= 11, prime
1 1+1+ 11= 13, prime
2 4+ 2+ 11= 17, prime
3 9+ 3+ 11= 23, prime
4 16+ 4+ 11= 31, prime
5 25+ 5+ 11= 41, prime
6 36+ 6+ 11= 53, prime
7 49+ 7+ 11= 67, prime
8 64+ 8+ 11= 83, prime
9 81+ 9+ 11= 101, prime
10 100+ 10+ 11= 121= 11*11! NOT PRIME!
If I were as smart as I like to think I am, I would have seen that "11" at the end and thought "11 is itself prime- the only way to factor 11 is "11*1" so immediately try 11^2+ 11+ 11= 11(11+ 1+ 1)= 11(13).
Another answer is
11 121+ 11+ 11= 143= 11*13 which I should have been able to get right away!
It is interesting that, after all those prime numbers, we get two non-primes in succession. What about 12? 144+ 12+ 11= 167 which is prime.