I need help solving a problem, I am stuck on this one.

Write 128 in each one of the following bases a.Five
b. Two
c. Twelve

a) 128 = (5 * 25) + (3 * 1) = (5 * 5^2) + (0 * 5^1) + (3 * 5^0)

in base 5 __ 503

b)128 = 2^7 = (1 * 2^7) + (0 * 2^6) + (0 * 2^5) + (0 * 2^4) + (0 * 2^3) + (0 * 2^2) + (0 * 2^1) + (0 * 2^0)
in base 2 __ 10000000

c) 128 = (10 * 12) + (8 * 1) = (10 * 12^1) + (8 * 12^0)
in base 12 __ A8

9 is our largest numerical digit, so we switch to letters after that

Scott's answer is correct except for A. 128 in base 5 should be written as 1003 because you cannot use the number 5 in your answer.

Sure, I can help you solve this problem!

a. To write 128 in base five, we need to find the highest power of five that is smaller than 128. In this case, it is 125 (5^3).

First, divide 128 by 125. The quotient is 1, and the remainder is 3. So the first digit (from left to right) in base five is 1.

Next, we need to find the highest power of five that is smaller than the remainder, which is 3. In this case, it is 25 (5^2).

Divide the remainder, 3, by 25. Since 3 is smaller than 25, the quotient is 0 and the remainder is 3. So the second digit in base five is 0.

Finally, the last digit is the remainder itself, which is 3 in this case.

Therefore, 128 in base five is written as 103.

b. To write 128 in base two (binary), we need to find the highest power of two that is smaller than 128. In this case, it is 128 itself (2^7).

First, divide 128 by 128. The quotient is 1, and the remainder is 0. So the first digit in base two is 1.

Next, we need to find the highest power of two that is smaller than the remainder, which is 0. In this case, it is 1 (2^0).

Divide the remainder, 0, by 1. Since 0 is smaller than 1, the quotient is 0 and the remainder is 0. So the second digit in base two is also 0.

Therefore, 128 in base two is written as 10000000.

c. Now, to write 128 in base twelve, we need to find the highest power of twelve that is smaller than 128. In this case, it is 144 (12^2).

First, divide 128 by 144. Since 128 is smaller than 144, the quotient is 0 and the remainder is 128. So the first digit in base twelve is 0.

Next, we need to find the highest power of twelve that is smaller than the remainder, which is 128. In this case, it is 12 (12^1).

Divide the remainder, 128, by 12. The quotient is 10, and the remainder is 8. So the second digit in base twelve is 10, which is represented by 'A' in this notation.

Finally, the last digit is the remainder itself, which is 8 in this case.

Therefore, 128 in base twelve is written as 0A8.