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- A historian analyzes a 2,400-year-old calendar system that cycles every 140 years. If a significant celestial event occurred in year 58 of the first cycle, and another in year 110 of the 14th cycle, what is the absolute difference in years between the two events? (Note: 14th cycle starts at (14 – 1) × 140 = 1820, so year 1 = 0)
- First, find year number of event at year 58 of first cycle: since cycles are 140 years, and first cycle starts year 0, event at year 58.
- 14th cycle starts at 13 × 140 = 1820; so year 110 of 14th cycle is 1820 + 110 = <<1820+110=1930>>1930.
- Absolute difference: 1930 – 58 = <<1930-58=1872>>1,872 years.
- #### 1872A climatologist analyzing satellite data observes that a particular Arctic region has lost ice cover at an average rate of 13.1 billion tons per year over the past 32 years. If the total ice mass lost during this period is equivalent to 420 billion tons, and the loss has been accelerating uniformly by 0.4 billion tons more per year each year, how much ice was lost in the final year?
- This is a uniformly accelerating sequence where total loss = sum of an arithmetic series.
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$4 \times 3 \times 12 = 144$
Therefore, the desired probability is:
\frac{144}{256} = \frac{9}{16}
Wait — but this counts only sequences with one pair and two singles, and one option unused. But is that the only case with exactly one repeated word? Yes, under the condition exactly two of the chosen options are the same and the other two are each different — this implies exactly one repetition, two others distinct. So the multiset is indeed {A,A,B,C} with B ≠ C ≠ A.
But note: the problem says exactly two of the chosen options are the same — this could be misinterpreted as at least two are the same, but the phrasing exactly two suggests a single pair and the rest different — i.e., one word appears twice, the others appear once — which is standard in such puzzles.
Thus, our count is correct.
Hence, the probability is:
But wait: $144 \div 16 = 9$, $256 \div 16 = 16$, yes: $\frac{144}{256} = \frac{9}{16}$?
Wait: $144 / 16 = 9$, $256 / 16 = 16$, so yes, $\frac{144}{256} = \frac{9}{16}$? Actually:
$144 = 16 \times 9$, $256 = 16 \times 16$, yes — so $\frac{144}{256} = \frac{9}{16}$? No: $16 \times 9 = 144$, $16 \times 16 = 256$, so $\frac{144}{256} = \frac{9}{16}$? Wait: $\frac{144}{256} = \frac{144 \div 16}{256 \div 16} = \frac{9}{16}$ — yes, correct.
