Number System

The number must divide (50 - 2) = 48 and (74 - 2) = 72. The greatest common divisor of 48 and 72 is 24.

The number must divide (78-3)=75, (94-4)=90, and (110-5)=105. The HCF of 75, 90, and 105 is 15.

The remainder when dividing by n is largest when n is slightly larger than a divisor of the number. For 817,380: 817,380 is divisible by 4, 5, 6, 9. When dividing by 8, 817,380 = 8 * 102,172 + 4. Testing remainders: 817,380 mod 4 = 0, mod 5 = 0, mod 6 = 0, mod 8 = 4, mod 9 = 0. The remainder 4 is the largest.

3^1 = 3 (mod 5), 3^2 = 9 = 4 (mod 5), 3^3 = 27 = 2 (mod 5), 3^4 = 81 = 1 (mod 5). The cycle is 4. 98 / 4 leaves remainder 2. Thus 3^98 = 3^2 = 9 = 4 (mod 5).

Let numbers be 36a and 36b where a, b are coprime. Product = 36a * 36b = 12960. ab = 12960 / 1296 = 10. Pairs of coprime factors of 10 are (1, 10) and (2, 5). There are 2 such pairs.

Using the division method for HCF in reverse: Last divisor = 75, last quotient = 3. Number = 75*3 + 0 = 225. Previous divisor = 225, quotient = 1. Number = 225*1 + 75 = 300. Previous divisor = 300, quotient = 1. Number = 300*1 + 225 = 525. Previous divisor = 525, quotient = 3. Number = 525*3 + 300 = 1875. The numbers are 525 and 1875.

The numbers are 4k, 5k, 6k where k = HCF = 18. The numbers are 72, 90, 108. The LCM of 72, 90, 108 is 1080.

We need the HCF of (211-4), (270-9), and (321-6), which are 207, 261, and 315. The HCF of 207, 261, and 315 is 9.

Let the numbers be 6a and 6b where a and b are coprime. LCM = 6ab = 60, so ab = 10. Pairs (a,b) are (1,10) and (2,5). Sum = 6(a+b) = 66, so a+b = 11. The pair (1,10) gives sum 11. Numbers are 6*1 = 6 and 6*10 = 60.

1001 sevens is 777...7 (1001 times). 1001 = 7 * 11 * 13. Since 1001 is a multiple of 7, 11, and 13, and the number consists of 1001 sevens, we can write it as 7 * (111...1). Dividing by 1001 is equivalent to dividing by 7 * 143. The remainder is 700.

The number N satisfies N = 7a + 6, a = 11b + 10, b = 13c + 12. Substituting back: a = 11(13c + 12) + 10 = 143c + 142. N = 7(143c + 142) + 6 = 1001c + 994 + 6 = 1001c + 1000. The remainder when divided by 1001 is 1000.

We need a number N such that N = 15k + 12 and N = 8m + 5. This implies N + 3 is divisible by 15 and 8. The LCM of 15 and 8 is 120. So N + 3 = 120k, or N = 120k - 3. For the largest 4-digit number, 120k - 3 <= 9999, so 120k <= 10002, k = 83. N = 120 * 83 - 3 = 9960 - 3 = 9957.

The number must be of the form LCM(4,5,6,7)*k + 3. LCM(4,5,6,7) = 420. Smallest 4-digit number of this form is 420*3 + 3 = 1263. Largest 3-digit number of this form is 420*2 + 3 = 843. Difference = 1263 - 843 = 420.

Number n = 6k + 1 = 7m + 2. Testing 961: 961/6 = 160 R 1 (True). 961/7 = 137 R 2 (True). This is the largest 3-digit number satisfying the conditions.