Number System

A number leaving remainder 1 when divided by both 5 and 6 must be of form 30k + 1 (LCM of 5 and 6 is 30). For two-digit numbers: k=0 gives 1 (not two-digit), k=1 gives 31, k=2 gives 61, k=3 gives 91. Checking: 91 ÷ 5 = 18 remainder 1, 91 ÷ 6 = 15 remainder 1. 91 is the largest such two-digit number.

To find the greatest number dividing 926 and 2313 leaving remainders 2 and 3 respectively, first subtract the remainders from the numbers. This gives us 924 (926-2) and 2310 (2313-3). Now find HCF of 924 and 2310. 924 = 2² × 3 × 7 × 11 = 462 × 2. 2310 = 2 × 3 × 5 × 7 × 11 = 462 × 5. HCF is 462. Verify: 926 ÷ 462 = 2 remainder 2, 2313 ÷ 462 = 5 remainder 3.

We need the largest 2-digit number N where N≡2(mod 3), N≡3(mod 4), N≡5(mod 6). Numbers ≡5(mod 6) are: 95, 89, 83, 77... Checking conditions: 95÷3=32r2✓, 95÷4=23r3✓, 95÷6=15r5✓. Sum of digits of 95 = 9+5=14. Note: 98 fails as 98÷4=24r2.

If x gives remainder 10 when divided by 15, 18, 20, 27, then x - 10 is divisible by all these numbers. LCM of 15, 18, 20, 27 = 540. So x - 10 = 540k. For least number, k = 1, giving x = 550. Option A matches.

LCM of 15, 18, 24, 45 = 360. Number = 360k + 4 (since remainder is 4). For largest 4-digit: 9999 ÷ 360 = 27.77, so k = 27. Number = 360 × 27 + 4 = 9724.

For 54321A to be divisible by 9, sum of digits must be divisible by 9. 5+4+3+2+1+A = 15+A must be divisible by 9, so A = 3 (since 15+3=18). For 8B2043 to be divisible by 19, testing values: if B=5, 852043 ÷ 19 = 44855 (exact division, remainder 0). Therefore B=5. When B is divided by A: 5 ÷ 3 = 1 with remainder 1, so remainder is 1.

A number leaving remainder 3 when divided by 9, 10, 12, and 15 must be of form LCM(9,10,12,15)k + 3. LCM(9,10,12,15) = 2² × 3² × 5 = 180. The smallest such number is 180×1 + 3 = 183. Distractor 177 = 180-3 (wrong sign) and 199, 218 don't fit the pattern.

Let the numbers be 5x and 7x. Their LCM is 35x = 315, so x = 9. The numbers are 45 and 63. Their product is 45 × 63 = 2835. Alternatively, for coprime ratios a:b, product = LCM × GCD. Since 5 and 7 are coprime, GCD = x = 9, so product = 315 × 9 = 2835.

Find LCM of 24, 60, 75, 90 = 1800. Greatest 5-digit number is 99999. 99999 ÷ 1800 = 55 remainder 999. 999 - 4 = 995. 99999 - 995 = 99004. Check: 99004 ÷ 24 = 4125 R4, ÷ 60 = 1650 R4, ÷ 75 = 1320 R4, ÷ 90 = 1100 R4.

Let number N = d × q + 8, where d is divisor. Then 2N = 2d × q + 16. When 2N is divided by d, remainder is 16 mod d. Given this remainder is 5, so 16 ≡ 5 (mod d), meaning 16 - 5 = 11 is divisible by d. Since 8 < d, d = 11.

For two numbers, the product of LCM and HCF equals the product of the numbers. Given LCM = 120, HCF = 3, and one number = 24, we have: 120 × 3 = 24 × x. Solving: 360 = 24x, so x = 15. The second number is 15.

To find the greatest number dividing 390, 495, and 300 without remainder, we need the HCF. Factorizing: 390 = 2 × 3 × 5 × 13, 495 = 3 × 3 × 5 × 11, 300 = 2 × 2 × 3 × 5 × 5. The common factors are 3 and 5, so HCF = 3 × 5 = 15. 5 is too small (common but not greatest). 25 and 35 are not factors of all three numbers.

If x gives the same remainder y for all three numbers, then x must divide the differences. 277 - 152 = 125, 427 - 277 = 150. GCD(125, 150) = 25. Testing: 152 = 25×6 + 2, 277 = 25×11 + 2, 427 = 25×17 + 2. So x = 25, y = 2. Therefore x - y = 23.

We need N such that N ≡ 34 (mod 40) and N ≡ 34 (mod 52), with N being the largest 4-digit number satisfying this. LCM(40, 52) = 520. Numbers of form 520k + 34 give remainder 34 when divided by both. Largest 4-digit: try k=19 gives 9914. Check: 9914 ÷ 40 = 247 rem 34, 9914 ÷ 52 = 190 rem 34. Correct.

To find the greatest number that divides 3050 and 5200 leaving remainders 7 and 9 respectively, we first subtract the remainders: 3050 - 7 = 3043, and 5200 - 9 = 5191. We need to find the HCF of 3043 and 5191. Factorizing: 3043 = 17 × 179, and 5191 = 29 × 179. The HCF is 179, which is the greatest such divisor.