The remainder when 123 is divided by 10 is 3, as 123 = 10*12 + 3.

When 37 is divided by 5, it goes 7 times (7*5=35) with a remainder of 2 (37-35=2).

Let the score in the third subject be x. The average is (90 + 80 + x) / 3 = 85. Solving gives x = 90.

The probability of losing all 5 games is (1 - 0.3)^5 = 0.168. Therefore, the probability of winning at least once is 1 - 0.168 = 0.832, which rounds to 0.836.

If A is a subset of B, it means that every element of A is contained within B.

The relationship of direct proportionality is represented by the equation x = ky, where k is a constant.

Total parts = 2 + 3 = 5. Y = (3/5) * 50 = 30 liters.

Total parts = 5 + 3 = 8. Volume of milk = (5/8) * 32 = 20 liters.

If milk is 5 parts, then water is 3 parts. Total parts = 5 + 3 = 8. Water = (3/5) * 40 = 24 liters.

Total parts = 5 + 3 = 8. Volume of milk = (5/8) * 64 = 40 liters.

Initial sugar = 1 part, water = 4 parts. Total = 5 parts. New sugar = 2 liters, water = 8 liters. Ratio = 2:8 = 1:4.

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

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

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.

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.

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

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

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

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.

'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.

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