'132' in base 4 is calculated as 1*4^2 + 3*4^1 + 2*4^0 = 16 + 12 + 2 = 30.

The sum of '12' and '21' in base-4 is '30' in base-4.

In base-4, the sum of 3 and 2 is 11 (3 + 2 = 5 in decimal, which is 11 in base-4).

To convert from base-5 to decimal, multiply each digit by 5 raised to the power of its position (from right to left, starting at 0). Thus, 2*5^2 + 4*5^1 + 3*5^0 = 2*25 + 4*5 + 3*1 = 50 + 20 + 3 = 73.

2*5^2 + 4*5^1 + 3*5^0 = 50 + 20 + 3 = 73.

In base-5, '23' is 2*5 + 3 = 13 and '14' is 1*5 + 4 = 9. Their sum is 22 in decimal, which is '42' in base-5.

In base-5, '34' is 3*5 + 4 = 19 and '21' is 2*5 + 1 = 11. Their sum is 30, which is '100' in base-5.

The base-5 number 243 is calculated as 2*5^2 + 4*5^1 + 3*5^0 = 50 + 20 + 3 = 60 in decimal.

The base-5 number 243 converts to decimal as 2*5^2 + 4*5^1 + 3*5^0 = 50 + 20 + 3 = 73.

In base-5, the decimal number 10 is represented as 20.

In base-5, the decimal number 12 is represented as 22 (2*5^1 + 2*5^0 = 12).

In base-5, the decimal number 7 is represented as 12, since 1*5^1 + 2*5^0 = 5 + 2 = 7.

In a base-5 numeral system, valid digits are 0, 1, 2, 3, and 4; thus, 5 is not valid.

'12' in base-6 is 1*6 + 2 = 8. '3' is 3. Their product is 8*3 = 24, which is '42' in base-6.

'24' in base-6 is 2*6 + 4 = 16 and '13' is 1*6 + 3 = 9. Their product is 16*9 = 144, which is '100' in base-6.

In base-6, '5' * '2' = 10 in decimal, which is '12' in base-6.

To convert from base-7 to decimal, calculate 3*7^2 + 4*7^1 + 5*7^0 = 3*49 + 4*7 + 5 = 147 + 28 + 5 = 180.

7*5 = 35 in decimal, which is '43' in base-8.

'110' (6 in decimal) + '101' (5 in decimal) = '1011' (11 in decimal).

Let the base be 'b'. Then, A3 = 10*b + 3 = 163. Solving gives b = 12.

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).

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.