Let the initial amount of sugar be x liters and water be 4x liters. After adding 2 liters of sugar, the new ratio becomes (x + 2) : 4x.

Initially, there is 1 part sugar and 4 parts water, totaling 5 parts. In 20 liters, there are 4 liters of sugar and 16 liters of water. After adding 2 liters of sugar, the new ratio is 6:16, which simplifies to 1:5.

Initial sugar = 1 liter, water = 16 liters. After adding 2 liters of sugar, the new ratio is 3:16, which simplifies to 1:5.

Initial sugar = 1 part, water = 4 parts (8 liters). After adding 2 liters of sugar, new sugar = 2 liters, water = 8 liters. Ratio = 2:8 = 1:4.

Water = 4 parts, Sugar = 1 part. Total parts = 5. Sugar = (1/4) * 20 = 4 liters.

If sugar:water = 1:5, then for 12 liters of sugar, water = 5 * 12 = 60 liters.

If sugar:water = 1:5, then for 12 liters of sugar, water = 5 * 12 = 60 liters.

Let the initial amounts be 2x and 3x. After adding 5 liters to the first component, the new ratio becomes (2x + 5):3x. Solving gives 3:2.

In modular arithmetic, if a ≡ b (mod m), then (a - b) is divisible by m. Here, 7 - 3 = 4, which is divisible by 4.

Since 7 - 3 = 4, n must be a divisor of 4. Therefore, n can be 1, 2, or 4, but for the congruence to hold, n must be greater than 4.

To convert from base 5 to decimal, calculate: 2*5^2 + 4*5^1 + 3*5^0 = 2*25 + 4*5 + 3*1 = 50 + 20 + 3 = 73.

'101' in binary is 5, so '111' in binary is 7.

'101' in binary is 5 in decimal. '110' in binary is 6 in decimal.

In base 5, '123' = 1*25 + 2*5 + 3 = 27.

'210' in base 3 is 2*3^2 + 1*3^1 + 0*3^0 = 18 + 3 + 0 = 21. Therefore, '102' in base 3 is 1*3^2 + 0*3^1 + 2*3^0 = 9 + 0 + 2 = 11.

'B' represents 11 in the same system where 'A' is 10.

'AB' means A*base + B = 11. Since A=10, we have 10*base + B = 11, so B=1.

'B' (18) is a multiple of 'A' (12) since 18 can be expressed as 12 multiplied by 1.5.

'A' can be 1, 2, 3, 4, 6, or 12 (factors of 12) and 'B' can be 3, 6, 9, 12, etc. The only combination that fits is A=3 and B=12, which gives us 36.

'A' can be 3 or 4 (factors of 12), and 'B' can be 3, 6, or 9 (multiples of 3). The only combination that fits is 3 and 4, which gives us 24.

In this case, A must be a factor of 12 (1, 2, 3, 4, 6, 12) and B must be a multiple of 3 (3, 6, 9, 12). The pair (2, 6) satisfies both conditions.

The number 1234 is expressed in base 8, as indicated by the powers of 8 used in its representation.

If the original number is x, then x + 12 is a multiple of 5. This implies that x must be one more than a multiple of 5.

If we add 12 to 8, we get 20, which is a multiple of 5.

A number is divisible by 3 if the sum of its digits is divisible by 3.

'B2' in this system is 11*12^1 + 2*12^0 = 132 + 2 = 134 in decimal.

'210' in base 4 is calculated as 2*4^2 + 1*4^1 + 0*4^0 = 32 + 4 + 0 = 36.

'123' in base 3 = 1*3^2 + 2*3^1 + 3*3^0 = 27. '321' in base 3 = 3*3^2 + 2*3^1 + 1*3^0 = 81.

'23' in this system is 15. '32' would represent 22 in decimal.

15% of 300 = 0.15 * 300 = 45.