To evaluate the expression \( 15 + 23 + 44 + 55 + 10 \) in base 6, we first need to convert all numbers from base 6 to base 10, perform the addition, and then convert the result back to base 6.
Step 1: Convert each number from base 6 to base 10.
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For \( 15_6 \): \[ 1 \times 6^1 + 5 \times 6^0 = 6 + 5 = 11_{10} \]
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For \( 23_6 \): \[ 2 \times 6^1 + 3 \times 6^0 = 12 + 3 = 15_{10} \]
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For \( 44_6 \): \[ 4 \times 6^1 + 4 \times 6^0 = 24 + 4 = 28_{10} \]
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For \( 55_6 \): \[ 5 \times 6^1 + 5 \times 6^0 = 30 + 5 = 35_{10} \]
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For \( 10_6 \): \[ 1 \times 6^1 + 0 \times 6^0 = 6 + 0 = 6_{10} \]
Step 2: Add the base 10 equivalents: \[ 11 + 15 + 28 + 35 + 6 \]
Calculating step by step: \[ 11 + 15 = 26 \] \[ 26 + 28 = 54 \] \[ 54 + 35 = 89 \] \[ 89 + 6 = 95 \]
So, the total in base 10 is \( 95_{10} \).
Step 3: Convert \( 95_{10} \) back to base 6.
We do this by continuously dividing by 6 and recording the remainders:
- \( 95 \div 6 = 15 \) remainder \( 5 \)
- \( 15 \div 6 = 2 \) remainder \( 3 \)
- \( 2 \div 6 = 0 \) remainder \( 2 \)
Now we compile the remainders from the last division to the first:
- Starting from the last remainder to the first gives: \( 235 \)
Thus, \( 95_{10} = 235_6 \).
The final answer is: \[ \boxed{235_6} \]