Primes and Composites: Additional Practice (Set 1)

Three are prime; one factors into two primes close together in size.

$323 = 17 \times 19$

So $323$ is composite. It is odd with digit sum $8$, so the rules for $2$ and $3$ miss it, and you find $17$ by trial division. The numbers $223$, $227$, and $229$ are prime.

Check $31$ through $39$, dropping evens and multiples of $3$ and $5$.

$31 \to \text{prime}, \quad 37 \to \text{prime}$

The rest are composite: $33 = 3 \times 11$, $35 = 5 \times 7$, $39 = 3 \times 13$, and $32, 34, 36, 38$ are even. So there are two primes.

Screen the composites: $377 = 13 \times 29$, $381 = 3 \times 127$ (digit sum $12$), and $385 = 5 \times 77$.

$383: \text{ odd, digit sum } 14, \text{ not ending in } 0 \text{ or } 5, \quad 19 \times 19 = 361 < 383 < 400 = 20 \times 20$

Test primes up to $19$: $7, 11, 13, 17, 19$ each leave a remainder. So $383$ is prime.

A number has exactly three factors only when it is a prime times itself.

$49 = 7 \times 7 \;\Rightarrow\; \text{factors } 1, 7, 49$

That is exactly three. The others have more: $45$ has six, $50$ has six, and $48$ has ten.

Three are prime; trial-divide the fourth past the small primes.

$403 = 13 \times 31$

So $403$ is composite. It is odd with digit sum $7$, so the rules for $2$ and $3$ miss it, and $13$ is the first divisor that works. The numbers $233$, $239$, and $241$ are prime.

Both numbers must be prime and differ by $2$.

$179 \text{ and } 181 \to \text{both prime}, \quad 181 - 179 = 2$

So $179$ and $181$ are twin primes. The others contain composites: $143 = 11 \times 13$, $185 = 5 \times 37$, and $169 = 13 \times 13$.

We want the factor pair in which both factors are greater than $1$, which rules out the trivial pair $1 \times 377$. Factor $377$.

$377 = 13 \times 29$

Both factors exceed $1$, and $13 \le 29$, so the smaller factor $a$ is $13$.

List the primes in that range, screening with the quick rules.

$61, \; 67, \; 71, \; 73, \; 79$

Each of these five is prime. The composites between are $63 = 9 \times 7$, $65 = 5 \times 13$, $69 = 3 \times 23$, $75 = 3 \times 25$, $77 = 7 \times 11$, and every even number. So the count is five.

Screen the composites: $291 = 3 \times 97$ (digit sum $12$), $437 = 19 \times 23$, and $427 = 7 \times 61$.

$439: \text{ odd, digit sum } 16, \text{ not ending in } 0 \text{ or } 5, \quad 20 \times 20 = 400 < 439 < 441 = 21 \times 21$

Test primes up to $20$: $7, 11, 13, 17, 19$ each leave a remainder. So $439$ is prime.

Test the suspicious claim with a counterexample.

$9 = 3 \times 3 \;\Rightarrow\; 9 \text{ is odd but composite}$

So "every odd number is prime" is false. The other three statements are all true: $2$ is the only even prime, every larger prime is odd, and $1$ is neither.

Work upward from $101$. Note $102$ is even and $105$ ends in $5$.

$101: \text{ odd, digit sum } 2, \text{ not ending in } 0 \text{ or } 5, \quad 11 \times 11 = 121 > 101$

So only $2, 3, 5, 7$ need checking, and none divides $101$. So $101$ is prime, and it is the smallest prime above $100$.

Three are prime; one is a product of two primes in the twenties.

$667 = 23 \times 29$

So $667$ is composite. It is odd with digit sum $19$, so the rules for $2$ and $3$ miss it, and you find $23$ by trial division (since $25 \times 25 = 625 < 667 < 676 = 26 \times 26$). The numbers $257$, $263$, and $269$ are prime.