The factors of 12 are 1, 2, 3, 4, 6, and 12. Their sum is 28.

All pairs listed are multiples of 7 that sum to 56.

The two numbers are 12 and 8, as they satisfy both the sum (12 + 8 = 20) and the product (12 * 8 = 96).

The sum of the remainders is 4 + 4 = 8, and 8 divided by 7 gives a remainder of 1.

If one number is 20 (a multiple of 5), the other number would be 10, making it a valid option.

5 and 25 add up to 30, and 25 is a multiple of 5.

5 and 25 add up to 30, and 5 is a multiple of 5.

12 is a multiple of 6, and 12 + 18 = 30.

Both numbers can be expressed as 6k + 4. Their sum is 12k + 8, which gives a remainder of 2 when divided by 6.

For the sum to be a multiple of 10, at least one of the numbers must be even.

The HCF of 18 and 24 is 6, as it is the largest number that divides both.

Let the numbers be 2x and 3x. Then, LCM(2x, 3x) = 6x = 60, which gives x = 10. The numbers are 20 and 30, and their sum is 50.

x + y ≡ 2 + 1 ≡ 3 (mod 3), which is equivalent to 0, but since we are looking for the value, it is 0 mod 3.

Using the method of successive substitutions or the Chinese Remainder Theorem, the smallest solution is x = 26.

x can be expressed as 4 + 5k, where k is an integer. 19 is not congruent to 4 modulo 5.

10 mod 6 = 4, so x could be 10.

Using the method of successive substitutions or the Chinese Remainder Theorem, we find that the smallest positive integer solution is x = 25.

x + y ≡ 4 + 3 ≡ 7 (mod 7), which is equivalent to 0.

11 mod 7 = 4, so x could be 11.

The decimal number 5 is represented as 12 in base 3 (1*3^1 + 2*3^0 = 3 + 2 = 5).

In base 3, the decimal number 5 is represented as '12' (1*3^1 + 2*3^0 = 3 + 2 = 5).

'54' in base 6 = 5*6^1 + 4*6^0 = 30 + 4 = 34.

'54' in base 6 is calculated as 5*6 + 4 = 30 + 4 = 34 in decimal.

To convert from base 6 to decimal, calculate: 2*6^2 + 4*6^1 + 5*6^0 = 2*36 + 4*6 + 5*1 = 72 + 24 + 5 = 101.

To convert from base 8 to decimal, calculate: 5*8^1 + 7*8^0 = 5*8 + 7*1 = 40 + 7 = 47.

'57' in base 8 is calculated as 5*8^1 + 7*8^0 = 40 + 7 = 47.

'10' in base 3 is 3 and '11' is 4. Their product is 12, which is '110' in base 3.

The decimal number 5 is represented as 21 in base-3 (2*3^1 + 1*3^0 = 5).

'210' in base-3 is 2*3^2 + 1*3^1 + 0*3^0 = 18 + 3 + 0 = 21 in decimal.

To add in base-3, convert to decimal: 12 (base-3) = 3 + 2 = 5 (decimal) and 21 (base-3) = 2 + 0 = 2 (decimal). The sum is 5 + 2 = 7 (decimal), which is 21 in base-3.