Number System

Find LCM of 24, 30, 54 = 1080. The required number leaves remainder 5, so number = LCM × k + 5. For k=1: 1085. Check: 1085 ÷ 24 = 45R5, 1085 ÷ 30 = 36R5, 1085 ÷ 54 = 20R5. So 1085 is the least such number.

LCM of 3, 4, 5 is 60. We need a number that leaves remainder 2 when divided by each, so number = 60k + 2. For k=1: 62 (smallest positive). Check: 62÷3=20 remainder 2, 62÷4=15 remainder 2, 62÷5=12 remainder 2. Option A (22) = 60×0+22 gives wrong remainders. Option B (42) fails for divisor 4. Option D (122) is 60×2+2 but not smallest.

To find the greatest number that divides 130, 305, and 245 leaving remainders 2, 1, and 5 respectively, we first subtract the remainders from each number: 130-2=128, 305-1=304, 245-5=240. Then we find the HCF of 128, 304, and 240. The HCF of these three numbers is 16, which is the required answer. This works because 16 divides all three adjusted numbers exactly, so it will divide the originals with the specified remainders.

Required number divides (61-1), (72-2), (83-3) i.e., 60, 70, 80 exactly. HCF(60, 70, 80) = 10. Check: 60 = 10×6, 70 = 10×7, 80 = 10×8. HCF is 10. Option D is correct.

We need a number that leaves remainder 3 when divided by 12, 18, 21, 28. So (N-3) must be divisible by LCM(12,18,21,28). LCM(12,18,21,28) = LCM(2²×3, 2×3², 3×7, 2²×7) = 2²×3²×7 = 4×9×7 = 252. So N-3 = 252k, N = 252k + 3. For greatest 3-digit number: 252×3 + 3 = 756 + 3 = 759. 252×4 + 3 = 1011 (4 digits). So answer is 759.

When a number N is divided by 899 and leaves remainder 63, we can write N = 899k + 63 for some integer k. Since 899 = 29 × 31, we need to find 63 mod 29. 63 = 29 × 2 + 5, so the remainder when N is divided by 29 is 5. The key is that 899 is a multiple of 29, so we only need to find 63 mod 29.

LCM of 16, 18, 20, 25 = 3600. The number leaves remainder 4 when divided by these, so it is of the form 3600k + 4. We need this to be divisible by 7: 3600k + 4 ≡ 0 (mod 7). Since 3600 = 7×514 + 2, we have 3600 ≡ 2 (mod 7). So: 2k + 4 ≡ 0 (mod 7), giving 2k ≡ 3 (mod 7), so k ≡ 5 (mod 7). The smallest such k is 5, giving the number: 3600×5 + 4 = 18004, which is divisible by 7 (18004/7 = 2572).

Find LCM of 8, 12, 15, 20 = 120. The required number must be 2 more than a multiple of 120. Check options: 360 ÷ 120 = 3, so 360 + 2 = 362 (not in options). 242 = 2 × 120 + 2 = 242 ✓. Wait, let me recalculate: LCM(8,12,15,20) = 120. We need a number N such that N ≡ 2 (mod 8), N ≡ 2 (mod 12), N ≡ 2 (mod 15), N ≡ 2 (mod 20). This means N-2 must be divisible by LCM = 120. So N = 120k + 2. For k=1: N=122. 122 ÷ 8 = 15R2, 122 ÷ 12 = 10R2, 122 ÷ 15 = 8R2, 122 ÷ 20 = 6R2. Yes, 122 is correct.

The number N leaves remainders: N ≡ 7 (mod 8), N ≡ 2 (mod 5), N ≡ 3 (mod 4), N ≡ 1 (mod 3). Testing option C: 387÷8=48 R7✓, 387÷5=77 R2✓, 387÷4=96 R3✓, 387÷3=129 R1✓. The reverse of 387 is 783. Other options fail one or more remainder checks.

We need N such that N ≡ 1 (mod 2), N ≡ 2 (mod 3), N ≡ 3 (mod 4), N ≡ 4 (mod 5), N ≡ 5 (mod 6). Notice each remainder is one less than the divisor. This means N + 1 is divisible by 2, 3, 4, 5, and 6. So N + 1 = LCM(2,3,4,5,6) × k = 60k. Therefore N = 60k - 1. For k = 1, N = 59. Checking: 59 ÷ 2 = 29 R1 ✓, 59 ÷ 3 = 19 R2 ✓, 59 ÷ 4 = 14 R3 ✓, 59 ÷ 5 = 11 R4 ✓, 59 ÷ 6 = 9 R5 ✓.

For two numbers, product = HCF × LCM. Given one number = 117, HCF = 13, LCM = 1989, we get other number = (13 × 1989) / 117 = 221. This is a direct application of the HCF-LCM product formula.

If both numbers leave the same remainder r when divided by n, then n divides their difference: 34041 - 32506 = 1535. Factorizing 1535 = 5 × 307. Testing divisors: 1535 ÷ 307 = 5, 1535 ÷ 453 ≈ 3.39. Both numbers must be > r, where r = 32506 mod n. For n=307: 32506 ÷ 307 = 105 remainder 271, 34041 ÷ 307 = 110 remainder 271. Same remainder confirms n=307.

Let numbers be 15a and 15b (since HCF=15). Then LCM = 15ab = 180, so ab = 12. Also 15a + 15b = 105, so a + b = 7. Factor pairs of 12: (1,12)→sum13, (2,6)→sum8, (3,4)→sum7. So a=3, b=4. Numbers are 15×3=45 and 15×4=60. Sum=105, HCF=15, LCM=180. Option D is correct.

Add 5 to each fraction: 3/4 becomes 8/9 (+0.111), 2/3 becomes 7/8 (+0.125), 4/7 becomes 9/12 (+0.071), 5/7 becomes 10/12 (+0.119), 6/7 becomes 11/12 (+0.048). Maximum change is for 2/3.

Let the numbers be 3x and 5x. Since 3 and 5 are coprime (no common factors), their LCM = 3 × 5 × x = 15x. Given LCM is 225, we have 15x = 225, so x = 15. The smaller number is 3 × 15 = 45. This works because when two numbers are coprime, the LCM is their product.