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Difference between revisions of "1960 AHSME Problems/Problem 33"

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

Latest revision as of 00:54, 26 July 2025

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 $\Rightarrow{\boxed{\textbf{(A)}}}$.

See Also

1960 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 32 		Followed by
Problem 34
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions
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