This question asks us to find the smallest positive number that is divisible by both 12 and 18.

Break down each number into its prime factors.

The LCM uses the highest power of each prime factor that appears.

Multiply the highest powers together.

The LCM of 12 and 18 is 36, which is the smallest number divisible by both numbers.

Numbers of form 7k+3: when k=5, number=7(5)+3=38.

Check: 38÷7=5 remainder 3 ✓.

Check others: 24÷7=3 rem 3 (close but let's verify 38 first), 38÷7 gives remainder 3 ✓

Let smaller odd number = x.

Then x+(x+2)=56.

So 2x+2=56, 2x=54, x=27.

Check: 27+29=56 ✓

9+8+7+6+5+4+3 = 42.

Then 4+2 = 6.

Wait, let me recalculate: Sum = 42, which reduces to 4+2=6.

The answer should be D.

Actually 9+8+7+6+5+4+3=42, 4+2=6.

Let numbers be 5x and 7x.

Then 5x + 7x = 120, so 12x = 120, x = 10.

Larger number = 7 × 10 = 70

Odd numbers between 10 and 30: 11, 13, 15, 17, 19, 21, 23, 25, 27, 29.

Count = 10

Let number be n.

Then n(n+1) = 342.

Solving: n² + n - 342 = 0.

Using quadratic formula or testing: 17 × 18 = 306 (no), 18 × 19 = 342 (yes).

So n = 18.

Wait, checking: 17 × 18 = 306, 18 × 19 = 342.

Answer is B.

Even numbers from 2 to 50: 2, 4, 6, ..., 50.

This is an AP with first term = 2, last term = 50, common difference = 2.

Number of terms = 25.

Sum = 25(2+50)/2 = 25 × 26 = 650

Let original number = x.

Then x + 25% of x = 500.

So x + 0.25x = 500, 1.25x = 500, x = 400

Let smaller number = x, larger = 2x.

Then 2x - x = 45, so x = 45.

Numbers are 45 and 90.