We need to find three consecutive integers that sum to 45, then identify the smallest one.

Step 1: Define the three consecutive integers

Let the smallest integer be \(x\). Then the three consecutive integers are:

Step 2: Set up the equation

The sum of these three integers equals 45:

Step 3: Simplify and solve

Combine like terms:

Subtract 3 from both sides:

Divide by 3:

Step 4: Verify

The three consecutive integers are 14, 15, and 16.

Answer: The smallest integer is \(x = 14\) (Option B)

Using Euclidean algorithm: 144 = 96×1 + 48, 96 = 48×2 + 0. Therefore HCF = 48. Alternatively, 144 = 2^4×3^2 and 96 = 2^5×3. HCF = 2^4×3 = 48.

We need LCM(6, 8). 6 = 2×3, 8 = 2³. LCM = 2³×3 = 24. Therefore, 24 is the smallest number divisible by both 6 and 8.

Using divisibility rule for 11: alternating sum of digits must be divisible by 11. For 121: (1-2+1) = 0, which is divisible by 11. Verification: 121 ÷ 11 = 11.

Smallest 4-digit number is 1000. For divisibility by 18, number must be divisible by both 2 and 9. 1000 ÷ 18 = 55.55... Next: 1008 ÷ 18 = 56. So 1008 is the answer.

Using the division algorithm: Number = (Divisor × Quotient) + Remainder. Number = (15 × 23) + 8 = 345 + 8 = 353.

Using Euclidean algorithm: 180 = 144×1 + 36; 144 = 36×4 + 0. Therefore HCF = 36.

Digit sum = 9 + 8 + 7 + 6 + 5 + 4 = 39.

Let numbers be x and y. x + y = 50 and x - y = 10. Adding: 2x = 60, so x = 30.

6³ = 216. Thus 216 is a perfect cube.