Primes and Composites: Additional Practice (Set 2)

Test the composites first: $143 = 11 \times 13$, $169 = 13 \times 13$, and $133 = 7 \times 19$.

$149: \text{ odd, digit sum } 14, \text{ not ending in } 0 \text{ or } 5, \quad 12 \times 12 = 144 < 149 < 169 = 13 \times 13$

So test primes up to $12$: $7, 11$ both leave a remainder. So $149$ is prime.

Screen the composites: $247 = 13 \times 19$, $249 = 3 \times 83$ (digit sum $15$), and $253 = 11 \times 23$.

$251: \text{ odd, digit sum } 8, \text{ not ending in } 0 \text{ or } 5, \quad 15 \times 15 = 225 < 251 < 256 = 16 \times 16$

Test primes up to $15$: $7, 11, 13$ each leave a remainder. So $251$ is prime.

Find the largest $d$ with $d \times d \le 127$.

$11 \times 11 = 121 \le 127, \qquad 12 \times 12 = 144 > 127$

So you test divisors up to $11$. Trying $2, 3, 5, 7, 11$ finds no factor, so $127$ is prime.

You try divisors $d$ while $d \times d \le 211$.

$14 \times 14 = 196 \le 211, \qquad 15 \times 15 = 225 > 211$

So the largest such $d$ is $14$. You then check the prime candidates up to $14$ ($2, 3, 5, 7, 11, 13$); none divides $211$, so $211$ is prime.

List the primes in that range, screening out evens and multiples of $3$ and $5$.

$41, \; 43, \; 47, \; 53, \; 59$

That is five primes. Numbers like $49 = 7 \times 7$ and $51 = 3 \times 17$ are composite and do not count.

Since $19 \times 19 = 361 < 391 < 400 = 20 \times 20$, you must test divisors up to $19$, so $17$ and $19$ still need checking.

$391 \div 17 = 23, \text{ so } 391 = 17 \times 23$

So $17$ divides $391$, making it composite. Stopping after $13$ would have been too early.

The smaller factor of a pair satisfies $a \times a \le n$. Find the factor pair of $143$.

$143 = 11 \times 13, \qquad 11 \times 11 = 121 \le 143 < 169 = 13 \times 13$

So the factor pair closest to balanced is $11 \times 13$, and the largest the smaller factor $a$ can be is $11$.

Screen the composites: $265$ ends in $5$, $261 = 3 \times 87$ (digit sum $9$), and $259 = 7 \times 37$.

$263: \text{ odd, digit sum } 11, \text{ not ending in } 0 \text{ or } 5, \quad 16 \times 16 = 256 < 263 < 289 = 17 \times 17$

Test primes up to $16$: $7, 11, 13$ each leave a remainder. So $263$ is prime.

Since $14 \times 14 = 196 < 221 < 225 = 15 \times 15$, you must test divisors up to $14$, so $13$ still needs checking.

$221 \div 13 = 17, \text{ so } 221 = 13 \times 17$

So $13$ divides $221$, making it composite. Stopping after $11$ would have been too early.

Check $101$ through $109$, dropping evens and multiples of $3$ and $5$.

$101, \; 103, \; 107, \; 109 \to \text{prime}$

The composites removed are $105 = 3 \times 35$ and the even numbers $102, 104, 106, 108$. That leaves four primes.

Both numbers must be prime and add to $60$.

$13 + 47 = 60, \text{ and both } 13 \text{ and } 47 \text{ are prime}$

So $13 + 47$ works. In the others, $30$, $6$, $54$, $45$, and $15$ are all composite.

Screen the composites: $361 = 19 \times 19$, $339 = 3 \times 113$ (digit sum $15$), and $391 = 17 \times 23$.

$293: \text{ odd, digit sum } 14, \text{ not ending in } 0 \text{ or } 5, \quad 17 \times 17 = 289 < 293 < 324 = 18 \times 18$

Test primes up to $17$: $7, 11, 13, 17$ each leave a remainder. So $293$ is prime.