一天中，時鐘的時針和分針會重合幾次？
How manytimes a day do the hands of a clock overlap?
答案：時針和分針每小時重疊一次，但在12小時內(nèi)會重疊11次，一天之內(nèi)重疊22次。這是因為在12時位置的指針重合已經(jīng)計算在內(nèi)。重合時間點分別是上午12:00，1:05，2:11，3:16，4:22，5:27，6:33，7:38，8:44，9:49，10:55以及下午12:00，1:05，2:11，3:16，4:22，5:27，6:33，7:38，8:44，9:49，10:55。
The handsoverlap once an hour, but 11 times in 12 hours and 22 times in a day. This isbecause the overlap at 12 has already been accounted for. The overlaps occur at12, 1.05, 2.11, 3:16, 4:22, 5:27, 6:33, 7:38, 8:44, 9:49 and 10:55 in themorning and after midday at 12, 1.05, 2:11, 3:16, 4:22, 5:27, 6:33, 7:38, 8:44,9:49 and 10:55.
再比如，應(yīng)聘者們曾經(jīng)回答過這樣一個令人啼笑皆非的難題：
全世界有多少位鋼琴調(diào)音師？
How manypiano tuners are there in the entire world?
這類謎題被稱為“費米問題”，命名來自物理學(xué)家恩里科·費米，他之所以聲名遠(yuǎn)揚(yáng)是因為他能夠在少量的給定信息甚至沒有信息的情況下進(jìn)行運算。費米問題意在考察應(yīng)聘者的估算能力以及量綱分析能力。
那么這個問題如何解答呢？
解決費米問題的方法在于通過一系列估算而無限接近正確答案。因此，應(yīng)聘者需要考量一些因素，諸如：擁有鋼琴的家庭戶數(shù)，此類家庭進(jìn)行鋼琴調(diào)音的頻次等，從而得出每年有多少次的鋼琴調(diào)音。隨后，應(yīng)聘者們需要估算出鋼琴調(diào)音師的平均工作時長以及工作量。
The puzzleis solved by multiplying a series of estimates to get to the right answer. So acandidate would have to estimate factors such as how many households have apiano, how often they are tuned to figure out how many piano tunings take placea year.They then need to calculate the average working hours of a piano tunersand the number of jobs they carry out.
因此，用每年所有家庭需要進(jìn)行鋼琴調(diào)音的次數(shù)除以每年每位鋼琴調(diào)音師的工作量，答案就此誕生。
The numberof piano tunings that take place per year divided by the number per year perpiano tuner should yield the answer.
