2001 CEMC Gauss (Grade 8) Problems/Problem 12

The following problem is from both the 2001 CEMC Gauss (Grade 8) #12 and 2001 CEMC Gauss (Grade 7) #15, so both problems redirect to this page.

Problem

A prime number is called a “Superprime” if doubling it, and then subtracting $1$ , results in another prime number. The number of Superprimes less than $15$ is

$\text{ (A) }\ 2 \qquad\text{ (B) }\ 3 \qquad\text{ (C) }\ 4 \qquad\text{ (D) }\ 5 \qquad\text{ (E) }\ 6$

Solution 1

First, we can list all of the prime numbers that are less than $15$ . They are $2$ , $3$ , $5$ , $7$ , $11$ , and $13$ . Now, we can test if these are super primes.

$2 \times 2 - 1 = 3$ , which is prime, so $2$ is a superprime.

$3 \times 2 - 1 = 5$ , which is prime, so $3$ is a superprime.

$5 \times 2 - 1 = 9$ , which is composite due to being divisible by $3$ , so $5$ is not a superprime.

$7 \times 2 - 1 = 13$ , which is prime, so $7$ is a superprime.

$11 \times 2 - 1 = 21 = 3 \times 7$ , which is composite, so $11$ is not a superprime

$13 \times 2 - 1 = 25 = 5^2$ , which is composite, so $13$ is not a superprime.

We see that of the primes in the list, there were $\boxed {\textbf {(B) } 3}$ superprimes.

~anabel.disher

2001 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
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CEMC Gauss (Grade 8)

2001 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources)
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CEMC Gauss (Grade 7)

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2001 CEMC Gauss (Grade 8) Problems/Problem 12

Problem

Solution 1