Question 239858: I came across this problem and had a mind blank on how to figure it out. But then as I read it, I began to think that it was just one of those questions that was suppose to seem like it was sort of hard, yet not, all the same. But here is the question:
"In China each calender year is given one of 12 names, which rotate year after year. The year 2000 was the year of the Dragon. The year 2001 was the year of the Snake. The subsequent ten years are, in order, the years of the Horse, Sheep, Monkey, Rooster, Dog, Bear, Rat, Ox, Tiger, and Rabbit. After the year of the Rabbit, the year of the Dragon will occur again, and then the whole cycle will repeat. What will the year 3000 be?"
Won't the year 3000 be the same as the year 2000 which was the Dragon if it's repeating in the same order every year? I feel like it's a trick question, yet it's not, but if anyone actually knows how to figure it out, please share! Much appreciated!
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! In China each calender year is given one of 12 names, which rotate year after year. The year 2000 was the year of the Dragon. The year 2001 was the year of the Snake. The subsequent ten years are, in order, the years of the Horse, Sheep, Monkey, Rooster, Dog, Bear, Rat, Ox, Tiger, and Rabbit. After the year of the Rabbit, the year of the Dragon will occur again, and then the whole cycle will repeat. What will the year 3000 be?"
Won't the year 3000 be the same as the year 2000 which was the Dragon if it's repeating in the same order every year? I feel like it's a trick question, yet it's not, but if anyone actually knows how to figure it out, please share!
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The animals repeat on a 12 year cycle. The year 3000 is 1000 years difference, which is 12*83 + 4.
2996 will again be the Dragon. 3000 will be 4 from that, which is the Monkey.
It's modulo arithmetic:
1000 mod 12 = 4, meaing 4 is the remainder
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