NUMBER

$10 = 3+7 = 5+5$. $16 = 3+13 = 5+11$. $24 = 5+19 = 11+13$. (Any one valid pair per number is sufficient.)

9 is odd, and the sum of two primes is even unless one of the primes is 2 (the only even prime). $9=2+7$ works using 2, but generally odd numbers greater than 2 cannot always be written this way, which is why the conjecture is specifically about even numbers — adding two odd primes always gives an even result.

Based on the pattern holding for every even number tested (and it has been verified by computer for extremely large numbers), the conjecture appears very likely to be true, even though it has never been formally proven for all even numbers.

A prime number has no factors other than 1 and itself. To prove a number is prime, every possible factor up to its square root must be checked — not just the first few primes. Since $\sqrt{91}\approx9.54$, factors up to 9 (i.e. 2, 3, 5, 7) all need checking.

Testing 7: $91 \div 7 = 13$ exactly. So $91 = 7 \times 13$, meaning 91 is composite, not prime — the student's claim is incorrect.

$$273 = 3 \times 7 \times 13$$

From the prime factors $3, 7, 13$, the possible factors of 273 (other than 1 and 273) are: $3, 7, 13, 21 (3\times7), 39 (3\times13), 91 (7\times13)$. Any of these could be the number of boxes per crate.

Only options of 25 or fewer boxes per crate work: 3, 7, 13, or 21 boxes per crate. E.g. with 21 boxes per crate, $273\div21=13$ crates would be needed.

$(3,5), (5,7), (11,13), (17,19), (29,31)$ — noting 31 is just over 30, so if strictly BOTH must be under 30: $(3,5),(5,7),(11,13),(17,19)$.

$(3,5)\to4$. $(5,7)\to6$. $(11,13)\to12$. $(17,19)\to18$. Every middle number (4,6,12,18) is EVEN.

$$\frac{p+(p+2)}{2}=p+1$$

If $p$ is odd, then $p+1$ is the number immediately after an odd number, which is always EVEN (odd numbers and even numbers strictly alternate on the number line) — so $p+1$ is guaranteed to be even whenever $p$ itself is odd.

Since $p+1$ is always even (for $p>2$), and the only even prime number is 2 itself, any even number greater than 2 must be composite (divisible by 2 and something else). So the middle number of a twin prime pair (other than the 2,3 pair) can never be prime.

A prime number is defined as a number with EXACTLY two distinct factors: 1 and itself. The number 1 has only ONE factor (itself, since $1\div1=1$ and there's no separate '1 and itself' distinction) — it fails the 'exactly two factors' requirement, so by definition 1 is neither prime nor composite; it's a special case of its own.

If 1 were considered prime, prime factorizations would no longer be unique — e.g. $12=2^2\times3$ could also be written as $12=1\times2^2\times3$, or $1\times1\times2^2\times3$, and so on infinitely, since multiplying by 1 never changes the value. Excluding 1 preserves the important property that every whole number greater than 1 has exactly ONE prime factorization.

$$91 = 7 \times 13$$

$221 = 13\times17$. As numbers get larger, the NUMBER of possible prime factors to test grows enormously (up to $\sqrt{n}$ for a number $n$), and there's no known fast shortcut — for genuinely huge numbers (hundreds of digits, as used in real encryption), even powerful computers would take an impractically long time to find the factors by testing, which is exactly the security principle real encryption systems rely on.

$(3,5), (5,7), (11,13), (17,19), (29,31)$ — noting 31 is just over 30, so if strictly BOTH must be under 30: $(3,5),(5,7),(11,13),(17,19)$.

$(3,5)\to4$. $(5,7)\to6$. $(11,13)\to12$. $(17,19)\to18$. Every middle number (4,6,12,18) is EVEN.

$$\frac{p+(p+2)}{2}=p+1$$

If $p$ is odd, then $p+1$ is the number immediately after an odd number, which is always EVEN (odd numbers and even numbers strictly alternate on the number line) — so $p+1$ is guaranteed to be even whenever $p$ itself is odd.

Since $p+1$ is always even (for $p>2$), and the only even prime number is 2 itself, any even number greater than 2 must be composite (divisible by 2 and something else). So the middle number of a twin prime pair (other than the 2,3 pair) can never be prime.

A prime number is defined as a number with EXACTLY two distinct factors: 1 and itself. The number 1 has only ONE factor (itself, since $1\div1=1$ and there's no separate '1 and itself' distinction) — it fails the 'exactly two factors' requirement, so by definition 1 is neither prime nor composite; it's a special case of its own.

If 1 were considered prime, prime factorizations would no longer be unique — e.g. $12=2^2\times3$ could also be written as $12=1\times2^2\times3$, or $1\times1\times2^2\times3$, and so on infinitely, since multiplying by 1 never changes the value. Excluding 1 preserves the important property that every whole number greater than 1 has exactly ONE prime factorization.

$$91 = 7 \times 13$$

$221 = 13\times17$. As numbers get larger, the NUMBER of possible prime factors to test grows enormously (up to $\sqrt{n}$ for a number $n$), and there's no known fast shortcut — for genuinely huge numbers (hundreds of digits, as used in real encryption), even powerful computers would take an impractically long time to find the factors by testing, which is exactly the security principle real encryption systems rely on.