B10 - Number Systems 1. Among the three numbers provided below, written in different number systems, find the maximum

, written in different number systems, find the maximum and write it as a decimal in the answer. Only write the number in the answer, without indicating the base of the number system. The numbers are: 10110100, 134, 583 Answer:

To find the maximum number among the given numbers, we need to convert them all into decimal form and compare their values.

Let"s start with the first set of numbers: 2316, 328, 111102.

To convert 2316 to decimal, we can use the formula:
\(2 \times 3^3 + 3 \times 3^2 + 1 \times 3^1 + 6 \times 3^0\).

This simplifies to:
\(2 \times 27 + 3 \times 9 + 1 \times 3 + 6 \times 1 = 54 + 27 + 3 + 6 = 90\).

So, 2316 in decimal form is 90.

Next, let"s convert 328 to decimal:
\(3 \times 8^2 + 2 \times 8^1 + 8 \times 8^0\).
This simplifies to:
\(3 \times 64 + 2 \times 8 + 8 \times 1 = 192 + 16 + 8 = 216\).

Therefore, 328 in decimal form is 216.

Now, let"s convert 111102 to decimal:
\(1 \times 2^5 + 1 \times 2^4 + 1 \times 2^3 + 1 \times 2^2 + 0 \times 2^1 + 2 \times 2^0\).
This simplifies to:
\(1 \times 32 + 1 \times 16 + 1 \times 8 + 1 \times 4 + 0 \times 2 + 2 \times 1 = 32 + 16 + 8 + 4 + 0 + 2 = 62\).

So, 111102 in decimal form is 62.

Comparing the decimal values, we can see that the maximum number among 90, 216, and 62 is 216. Therefore, the answer is 216.

Moving on to the second set of numbers: 2A16, 448, 1001112.

To convert 2A16 to decimal, we can break it down into \(2 \times 16^1 + A \times 16^0\).

Since A represents 10 in hexadecimal, the equation becomes:
\(2 \times 16^1 + 10 \times 16^0\).

Evaluating this, we have:
\(2 \times 16 + 10 \times 1 = 32 + 10 = 42\).

Thus, 2A16 in decimal form is 42.

Next, let"s convert 448 to decimal, which is simple as it is already in decimal form.

Therefore, 448 in decimal form is 448.

Now, let"s convert 1001112 to decimal:
\(1 \times 2^6 + 0 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 2 \times 2^0\).

Simplifying this, we get:
\(1 \times 64 + 0 \times 32 + 0 \times 16 + 1 \times 8 + 1 \times 4 + 1 \times 2 + 2 \times 1 = 64 + 8 + 4 + 2 + 2 = 80\).

So, 1001112 in decimal form is 80.

Comparing the decimal values, we can see that the maximum number among 42, 448, and 80 is 448. Therefore, the answer is 448.

Now we move on to the third set of numbers: 10110100, 134, 583.

To convert 10110100 to decimal, we can apply the formula:
\(1 \times 2^7 + 0 \times 2^6 + 1 \times 2^5 + 1 \times 2^4 + 0 \times 2^3 + 1 \times 2^2 + 0 \times 2^1 + 0 \times 2^0\).

Simplifying this, we get:
\(1 \times 128 + 0 \times 64 + 1 \times 32 + 1 \times 16 + 0 \times 8 + 1 \times 4 + 0 \times 2 + 0 \times 1 = 128 + 32 + 16 + 4 = 180\).

So, 10110100 in decimal form is 180.

Next, let"s convert 134 to decimal, which is already in decimal form.

Therefore, 134 in decimal form is 134.

Lastly, let"s convert 583 to decimal, also in decimal form.

Therefore, 583 in decimal form is 583.

Comparing the decimal values, we can see that the maximum number among 180, 134, and 583 is 583. Therefore, the answer is 583.

To summarize the answers:
1. The maximum among 2316, 328, and 111102 is 216 in decimal form.
2. The maximum among 2A16, 448, and 1001112 is 448 in decimal form.
3. The maximum among 10110100, 134, and 583 is 583 in decimal form.

I hope this detailed explanation helps you understand how to convert numbers between different number systems and find the maximum among them! If you have any further questions, feel free to ask!