Govt. Exams
Entrance Exams
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 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.
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 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 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)
We need a number of the form 8k + 5. Check: 29 = 8(3) + 5 = 24 + 5. ✓
We need to find x where x² = 169. Taking square root: x = √169 = 13.
12 = 2² × 3, 18 = 2 × 3², 24 = 2³ × 3. LCM = 2³ × 3² = 8 × 9 = 72.
Let the number be x. According to the problem: 5x - 3 = 47. Therefore, 5x = 50, so x = 10.
For divisibility by 6, number must be divisible by both 2 and 3. 104÷6 = 17.33... (not divisible). Others: 72, 84, 90 are all divisible by 6
