+ 2025 + 2027 + 2029 = (2023 + 2029) + (2025 + 2027) = 4052 + 4052 = 8104
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- 314_7 = 3 \cdot 7^2 + 1 \cdot 7^1 + 4 \cdot 7^0 = 3 \cdot 49 + 1 \cdot 7 + 4 = 147 + 7 + 4 = 158
- \boxed{158}
- Question: What is the remainder when $ 2023 + 2025 + 2027 + 2029 $ is divided by 8?
- Now compute $ 8104 \mod 8 $. Since 8100 is divisible by 8 (as $ 8100 \div 8 = 1012.5 $, but $ 8000 \div 8 = 1000 $, and $ 100 \div 8 = 12.5 $), we check only the last three digits:
- \div 8 = 13 \text{ exactly}, so } 8104 \equiv 0 \pmod{8}
- $$Question: How many ways are there to distribute 4 distinct chemical samples into 2 identical storage containers such that each container has at least one sample?
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