Numbers and Divisibility

Prime numbers are defined as numbers greater than 1 that are divisible only by 1 and themselves. The examples given (2, 3, 5, 7, 11) are all prime numbers as they each have exactly two factors.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. The sequence 2, 3, 5, 7, 11 consists of the smallest prime numbers.

Numbers divisible by 5 but not 2 are: 5, 15, 25, ..., 995 - an AP with 100 terms. Sum = 100/2 × (5 + 995) = 50000.

For a number to be divisible by 80, it must be divisible by both 5 and 16. Divisibility by 5 requires the last digit to be 0 or 5. Since xy is a two-digit number at the end, we need 65300 + 100x + 10y ≡ 0 (mod 80). This simplifies to 60 + 10x + y ≡ 0 (mod 80). Testing values, when x=6 and y=0, we get 60 + 60 + 0 = 120, which is divisible by 80. Therefore x+y=6+0=6.

We need N-3 to be divisible by 21, 28, 36, and 45. Therefore N-3 must equal the LCM of these numbers. LCM(21,28,36,45) = LCM(3×7, 2²×7, 2²×3², 3²×5) = 2² × 3² × 5 × 7 = 4 × 9 × 5 × 7 = 1260. So N = 1260 + 3 = 1263.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. Natural and whole numbers describe broader sets, while real numbers include non-integers.

A conditional 'If P, then Q' is false only when P is true and Q is false. If x = 32, it is divisible by 8 (true), but it is not divisible by 6 (false). For x = 48, both are true; for x = 6, P is false.

The statement 'divisible by 2 OR divisible by 3' is false only when BOTH conditions are false. We need a number not divisible by 2 (not even) AND not divisible by 3. Checking options: 6 is divisible by both (true), 8 is divisible by 2 (true), 15 is divisible by 3 (true), but 11 is not divisible by 2 (it's odd) and not divisible by 3 (1+1=2, not divisible by 3). Only 11 makes the OR statement false.

Using the Remainder Theorem, for a polynomial P(x) to be divisible by (x + 3), P(-3) must be 0. P(-3) = (-3)^3 + 4(-3)^2 - 7(-3) + 12 = -27 + 36 + 21 + 12 = 42. Thus, 42 must be subtracted to make the remainder zero.

A number is divisible by 3 if and only if the sum of its digits is divisible by 3. For example, 123 is divisible by 3 because 1+2+3=6, which is divisible by 3. This rule works for all positive integers.

To check divisibility by 11, calculate the alternating sum of digits. For 415624: 4-1+5-6+2-4 = 0. Since the result is 0 (which is divisible by 11), the number 415624 is divisible by 11. The other options fail this test.

Using the identity: if a + b = c, then a³ + b³ - c³ is divisible by a, b, and c. Here 21 + 33 = 54, so 54³ - 33³ - 21³ can be rearranged as 54³ - (33³ + 21³). However, note that for a + b - c = 0, the expression a³ + b³ + c³ - 3abc is divisible by (a + b + c). In this case: 54³ - 33³ - 21³ = 54³ + (-33)³ + (-21)³ - 3(54)(-33)(-21). Since 54 + (-33) + (-21) = 0, it's divisible by 0, meaning divisible by all three numbers individually.

1056 divided by 23 gives 45 remainder 21 (since 23 x 45 = 1035). To reach the next multiple of 23, we need to add (23-21) = 2 more. So 1056 + 2 = 1058, which is divisible by 23 (1058 / 23 = 46).