MathDB

17

Part of 2017 AMC 12/AHSME

Problems(2)

I think I spoiled this one

Source: 2017 AMC 12A #17

2/8/2017
There are 24 different complex numbers zz such that z24=1z^{24} = 1. For how many of these is z6z^6 a real number?
<spanclass=′latex−bold′>(A)</span>1<spanclass=′latex−bold′>(B)</span>3<spanclass=′latex−bold′>(C)</span>6<spanclass=′latex−bold′>(D)</span>12<spanclass=′latex−bold′>(E)</span>24<span class='latex-bold'>(A) </span>1\qquad<span class='latex-bold'>(B) </span>3\qquad<span class='latex-bold'>(C) </span>6\qquad<span class='latex-bold'>(D) </span>12\qquad<span class='latex-bold'>(E) </span>24
2017 AMC 12AAMC 12AMCcomplex numbers
Two Games

Source: 2017 AMC 12 B #17

2/16/2017
A coin is biased in such a way that on each toss the probability of heads is 23\frac{2}{3} and the probability of tails is 13\frac{1}{3}. The outcomes of the tosses are independent. A player has the choice of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?
<spanclass=′latex−bold′>(A)</span> The probability of winning Game A is 481 less than the probability of winning Game B.<span class='latex-bold'>(A)</span> \text{ The probability of winning Game A is }\frac{4}{81}\text{ less than the probability of winning Game B.} <spanclass=′latex−bold′>(B)</span> The probability of winning Game A is 281 less than the probability of winning Game B.<span class='latex-bold'>(B)</span> \text{ The probability of winning Game A is }\frac{2}{81}\text{ less than the probability of winning Game B.} <spanclass=′latex−bold′>(C)</span> The probabilities are the same.<span class='latex-bold'>(C)</span> \text{ The probabilities are the same.} <spanclass=′latex−bold′>(D)</span> The probability of winning Game A is 281 greater than the probability of winning Game B.<span class='latex-bold'>(D)</span> \text{ The probability of winning Game A is }\frac{2}{81}\text{ greater than the probability of winning Game B.} <spanclass=′latex−bold′>(E)</span> The probability of winning Game A is 481 greater than the probability of winning Game B.<span class='latex-bold'>(E)</span> \text{ The probability of winning Game A is }\frac{4}{81}\text{ greater than the probability of winning Game B.}
AMCAMC 12AMC 12 Bprobability2017 AMC 12B