1960 AHSME Problems/Problem 33

Revision as of 01:23, 13 May 2018 by Rockmanex3 (talk | contribs) (Solution)

Problem

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:

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 57\qquad \textbf{(E)}\ 58$

Solution

First, note that $n$ does not have a prime number larger than $61$ as one of its factors. Also, note that $n$ does not equal $1$.

Therefore, since the prime factorization of $n$ only has primes from $2$ to $59$, $n$ and $P$ share at least one common factor other than $1$. Therefore $P+n$ is not prime for any $n$, so the answer is $\boxed{\textbf{(A)}}$.

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 32
Followed by
Problem 34
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