# Counterexamples

#### mitchfel

##### New member
If n is a whole number, then n^2+ n + 11 is a prime number. Give me counterexamples of it

#### tkhunny

##### Moderator
Staff member
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?

#### mitchfel

##### New member
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

#### tkhunny

##### Moderator
Staff member
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
n * (n+1)

Consecutive Integers.

One of them is divisible by 11.

Maybe 10 and 11 or 11 and 12?
Maybe 21 and 22 or 22 and 23?
Maybe 32 and 33 or 33 and 34?

There are other types of counter examples, but this variety seems the most obvious.

#### mitchfel

##### New member
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?

#### mitchfel

##### New member
I get it now. Thank you so much

#### Subhotosh Khan

##### Super Moderator
Staff member
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?
If you select n = 10 or n = 21 or n = 32 what do get? Can you generalize this process (n = ?)

#### Jomo

##### Elite Member
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?
Wait wait, finally a problem that I can answer! How about 10 and 11 or even 11 and 12!!

#### HallsofIvy

##### Elite Member
If n is a whole number, then n^2+ n + 11 is a prime number. Give me counterexamples of it
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).
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.

• Subhotosh Khan

#### Subhotosh Khan

##### Super Moderator
Staff member
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).