This question tests integer properties and divisibility, examining conditions for a square to be divisible by 12. The key rule is that 12 = 4 * 3, so $t^2$ must be divisible by both 4 (requiring t even) and 3 (requiring t divisible by 3, as 3 is prime). Applying this, t even ensures $t^2$ ≡ 0 mod 4, and t divisible by 3 ensures $t^2$ ≡ 0 mod 3, combining to $t^2$ ≡ 0 mod 12. Choice D (t even and divisible by 3) must be true and satisfies all conditions, equivalent to t divisible by 6. For example, choice A assumes t divisible by 12, but t=6 is even and divisible by 3 with $t^2$=36 divisible by 12, yet 6 not divisible by 12. Similarly, choice E assumes t odd, but odd t yields odd $t^2$ not divisible by 4.

This question tests integer properties and divisibility, using modular arithmetic modulo 4. The key rule is that congruences determine parities: a ≡ 3 mod 4 means a odd, and b ≡ 2 mod 4 means b even. Applying this, a + b ≡ 3 + 2 ≡ 1 mod 4, which is odd, as odd + even = odd. No other choices are always true, such as a + b divisible by 4 (≡1 ≠0 mod 4). Thus, choice D (a + b is odd) is true and satisfies all conditions. For example, choice A assumes a + b divisible by 4, but ≡1 mod 4 contradicts this. Similarly, choice E assumes ab odd, but odd * even = even, failing the condition.

This question tests integer properties and divisibility, particularly the relationship between GCD and LCM. The key rule is that gcd(m, n) * lcm(m, n) = m * n, so here 6 * 180 = 1080 = m * n. Applying this, let m = 6a and n = 6b with gcd(a, b) = 1, then lcm(m, n) = 6 * a * b = 180, so a * b = 30. Possible pairs (a, b) with gcd 1 include (3, 10), giving m = 18 and n = 60, which satisfy gcd=6 and lcm=180. Thus, 18 could be the value of m and satisfies all conditions. For example, choice C (24) fails because 24 = 6*4, but 30/4 is not an integer, so no corresponding b with gcd(a, b)=1. Similarly, choice E (45) assumes a=7.5, which is not an integer.

This question tests integer properties and divisibility, using congruences modulo 6. The key rule is that x ≡ 1 mod 6 and y ≡ 5 mod 6 imply x = 6p + 1 and y = 6q + 5. Applying this, x + y = 6(p + q + 1), which is divisible by 6. Other properties like xy even do not always hold, as x=1 and y=5 give odd xy=5. Thus, choice A (x + y divisible by 6) must be true and satisfies all conditions. For example, choice D assumes x + y odd, but it is always even, contradicting the assumption. Similarly, choice E assumes xy even, but it can be odd when both are odd.

This question tests integer properties and divisibility, combining divisibility and parity. The key rule is that n divisible by 4 implies n even (actually n ≡ 0 mod 4), so n + 2k = 4m + 2k = 2(2m + k), which is even. Applying this, an even number cannot be odd, so no such n and k exist. Attempts to find examples always yield even n + 2k, contradicting the odd requirement. Thus, choice E (no such integers exist) must be true and satisfies all conditions. For example, choice A assumes k even, but even if k even, n + 2k remains even, failing the odd condition. Similarly, choice C assumes n odd, but n divisible by 4 cannot be odd.

This question tests integer properties and divisibility, focusing on multiples and common factors. The key rule is that if r is a multiple of 9 and s of 6, then r ≡ 0 mod 9 and s ≡ 0 mod 6, implying both are multiples of 3. Applying this, rs = (9k)(6m) = 54km, which is divisible by 3 (actually by 54). Other expressions like r + s = 9k + 6m = 3(3k + 2m) are also multiples of 3, but the question specifies rs. Thus, choice C (rs) must be a multiple of 3 and satisfies all conditions. For example, choice D (r/s) assumes a ratio that may not be an integer, let alone a multiple of 3. Similarly, choice A (r + s) is a multiple of 3 but is not the only one.

This question tests integer properties and divisibility, using congruences modulo 3. The key rule is that a ≡ 2 mod 3 and b ≡ 2 mod 3 imply a - b ≡ 0 mod 3, as both have the same remainder. Applying this, a - b = (3p + 2) - (3q + 2) = 3(p - q), divisible by 3. Other expressions like a + b ≡ 4 ≡ 1 mod 3 are not divisible by 3. Thus, choice C (a - b divisible by 3) must be true and satisfies all conditions. For example, choice A assumes a + b divisible by 3, but ≡1 mod 3 contradicts this. Similarly, choice D assumes remainder 2 for a + b, but it is actually 1 mod 3.

This question tests integer properties and divisibility, focusing on multiples and their products. The key rule is that a multiple of 4 and b multiple of 6 imply a = 4p and b = 6q, so ab = 24pq, divisible by 24. Applying this, the product always includes factors 83=24 (from 46). Other statements like a + b multiple of 12 do not always hold, as 4+6=10 not divisible by 12. Thus, choice B (ab multiple of 24) must be true and satisfies all conditions. For example, choice A assumes a + b multiple of 12, but fails for a=4, b=6 due to invalid assumption about additivity. Similarly, choice D assumes gcd=2, but can be larger like 12 for a=12, b=12.

This question tests integer properties and divisibility, focusing on powers of 2 when one factor is odd. The key rule is that if a is odd and ab divisible by $8=2^3$, then b must provide all three factors of 2, so b divisible by 8. Applying this, if b not divisible by 8, ab would have fewer than three factors of 2. Examples like a=1 (odd), b=8 give ab=8 divisible by 8. Thus, choice A (b divisible by 8) must be true and satisfies all conditions. For example, choice B assumes b even (divisible by 2), but b=2 gives ab=2a divisible by 2 not 8, an invalid assumption about insufficient powers. Similarly, choice D assumes divisible by 2 but not 4, yielding ab divisible by 2 not 8.

This question tests integer properties and divisibility, distinguishing multiples of 15 from those of 45. The key rule is that divisible by 15 means divisible by 3 and 5, but not by 45 means not divisible by 9 $(3^2$). Applying this, -30 is divisible by 15 (30/15=2) but not by 45 (30/45=2/3 not integer), with exactly one factor of 3. Zero is divisible by both, failing the 'not' condition. Thus, -30 could be the value of n and satisfies all conditions. For example, choice A (45) is divisible by 45, violating the condition. Similarly, choice E (0) is divisible by 45, as 0=45*0.