Entrance Exams
Govt. Exams
Digit sum = 9 + 8 + 7 + 6 + 5 + 4 = 39.
Since 9 and 11 are coprime (HCF=1), a number divisible by both must be divisible by their product: 9×11 = 99.
Using Euclidean algorithm: 180 = 144×1 + 36; 144 = 36×4 + 0. Therefore HCF = 36.
Let number = 10a + b. Reversed = 10b + a. Given: (10b + a) - (10a + b) = 45, so 9b - 9a = 45, thus b - a = 5. Also |a - b| = 5 or a - b = 5. From b - a = 5 and a - b could be -5 or 5. Testing: if b - a = 5 and digits sum conditions... Let a = 2, b = 7: number = 27. Reversed = 72. 72 - 27 = 45. ✓
Let the two consecutive even numbers be n and n+2. Then n(n+2) = 528. So n² + 2n - 528 = 0. Using quadratic formula or factoring: (n+24)(n-22) = 0. So n = 22 (taking positive value). The two numbers are 22 and 24. Larger = 24.
Perfect squares from 1 to 1000 are 1², 2², 3², ..., n² where n² ≤ 1000. So n ≤ √1000 ≈ 31.62. Therefore n can be 1, 2, 3, ..., 31. Total = 31 perfect squares.
A number is divisible by 9 if sum of its digits is divisible by 9. Here sum is 12, which is not divisible by 9. However, any number divisible by 9 is also divisible by 3. But the given condition states sum of digits is 12, and divisible by 9, which is contradictory. Re-reading: if divisible by 9, then sum must be divisible by 9. Since sum is 12 and divisible by 3, the number is divisible by 3.
Let number be n. n ≡ 4 (mod 7) and n ≡ 6 (mod 11). From first: n = 7k + 4. Substituting in second: 7k + 4 ≡ 6 (mod 11), so 7k ≡ 2 (mod 11). Testing values: k = 4 gives 7(4) + 4 = 32 ≡ 10 (mod 11). Try k = 5: 7(5) + 4 = 39 ≡ 6 (mod 11). Yes, 39 works.
Factoring: x² - 5x + 6 = (x - 2)(x - 3) = 0. Therefore x = 2 or x = 3.
Numbers divisible by both 4 and 6 are divisible by LCM(4,6) = 12. Numbers from 1 to 500 divisible by 12: ⌊500/12⌋ = 41.666..., so 41 numbers. Actually, ⌊500÷12⌋ = 41, but we need to check: 12 × 41 = 492. So there are 41 numbers. Let me recalculate: 500 ÷ 12 = 41.666, so answer is 41. Wait, the options suggest 42. Let me verify: counting from 12, 24, 36...492. That's 492/12 = 41. The correct count is 41, but closest option is 42.