### Instruction: You are given a sequence of $58$ terms; each term has the form $P+n$ where $P$ stands for the product $2 \times 3 \times 5 \times\ldots \times 61$ of all prime numbers less than or equal to $61$, and $n$ takes, successively, the values $2, 3, 4,\ldots, 59$. Let $N$ be the number of primes appearing in this sequence. Then $N$ is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 57\qquad \textbf{(E)}\ 58$.### Response: This is a problem from the AMC 10, a math contest for high school students.
I need to find how many of the terms in the sequence are prime numbers.
I notice that $P$ is a huge number that is divisible by every prime up to $61$.
So, adding any of those primes to $P$ will not make a prime number, since they will have a common factor.
That means I can eliminate any term in the sequence that has $n$ equal to a prime less than or equal to $61$.
There are $17$ such primes, so I can ignore those terms.
What about the other values of $n$?
They are all composite numbers, meaning they have at least two prime factors.
If I add a composite number to $P$, will I ever get a prime number?
I think about some examples.
Suppose $n$ is $6$, which is $2 \times 3$.
Then $P+6$ is divisible by $2$ and $3$, so it's not prime.
Suppose $n$ is $35$, which is $5 \times 7$.
Then $P+35$ is divisible by $5$ and $7$, so it's not prime.
I see a pattern.
If $n$ is composite, then it has some prime factor $p$ that is less than or equal to $n$.
But then $P$ is also divisible by $p$, since $p$ is less than or equal to $61$.
So, $P+n$ is divisible by $p$, and it's not prime.
This means that no term in the sequence can be prime, since either $n$ is prime and $P+n$ has a common factor with $P$, or $n$ is composite and $P+n$ has a common factor with $n$.
Therefore, the answer is $N = 0$.