Number System

We need a number N such that N = 8k + 3 and N = 9m + 1. This implies N is congruent to 3 mod 8 and 1 mod 9. Testing the largest 5-digit numbers near 99999, 99955 divided by 8 gives 12494 remainder 3, and 99955 divided by 9 gives 11106 remainder 1.

N must divide (50-2)=48, (75-3)=72, and (125-5)=120. The maximum value of N is the greatest common divisor (GCD) of 48, 72, and 120. The GCD(48, 72, 120) = 24.

The difference between divisor and remainder is constant (3-2=1, 4-3=1, ..., 10-9=1). The number is LCM(3, 4, 5, 6, 7, 8, 9, 10) - 1. LCM(3, 4, 5, 6, 7, 8, 9, 10) = 2520. 2520 - 1 = 2519.

n = 5k + 3. Then 8n = 8(5k + 3) = 40k + 24. 24 divided by 5 leaves a remainder of 4.

Successive division by 11, 12, 15 with remainders 7, 3, 10. Let the number be N. N = 15k + 10. Previous quotient Q1 = 12(15k+10) + 3 = 180k + 123. N = 11(180k + 123) + 7 = 1980k + 1353 + 7 = 1980k + 1360. For largest 4-digit number, 1980k + 1360 <= 9999. 1980k <= 8639. k=4. N = 1980(4) + 1360 = 7920 + 1360 = 9280.

LCM of 3, 4, 5, 6 is 60. Multiples of 60: 60, 120, ..., 960, 1020. The smallest 4-digit multiple is 1020. Adding the remainder 2 gives 1022.

To find the greatest 10-digit number with distinct digits divisible by 8, we arrange the digits in descending order as much as possible while ensuring the last three digits form a number divisible by 8. This gives the number 9876543120. Dividing this number by 11 yields a remainder of 4.

Let the divisor be D. N = qD + 21. 3N = q'D + 9. Since 3N = 3(qD + 21) = 3qD + 63, the remainder when 3N is divided by D is the remainder of 63 divided by D. 63 = kD + 9, so kD = 54. Possible divisors are factors of 54 greater than 21. 27 is a factor of 54 and greater than 21.

Find LCM of 12, 16, 18. 12 = 2^2 * 3, 16 = 2^4, 18 = 2 * 3^2. LCM = 2^4 * 3^2 = 16 * 9 = 144. Number = LCM + remainder = 144 + 5 = 149.

Let the numbers be 81a and 81b, where a and b are coprime. 81(a+b) = 1215, so a+b = 15. Pairs (a,b) such that gcd(a,b)=1 are (1,14), (2,13), (4,11), (7,8). There are 4 such pairs.

The Least Common Multiple (LCM) of two numbers must always be divisible by their Highest Common Factor (HCF). Since 82 is not divisible by 9, it cannot be the LCM of any two numbers whose HCF is 9.

The greatest number N that divides 66, 88, 110 leaving the same remainder is the HCF of (88-66) and (110-88), which is HCF(22, 22) = 22. Sum of digits of 22 is 2 + 2 = 4.