full transcript
From the Ted Talk by Alex Rosenthal: Can you solve the human cannonball riddle?
Unscramble the Blue Letters
Up to this point we haven’t used the rule that each chamber must have 1 to 3 cells. It tells us that the miuimnm for the lower level is 8 cells, in which case, by rule 1, the upper level would have 16. On every side, the lower level would account for 3 of the 11 cells, so the upper would have to have 8. But if two opposite sides had 8, that would be the upper level’s entire 16, not leaving any for these two chambers. So 24 is out.
Let’s look at the other extreme. The upper level can have at most 3 times 8, or 24 cells, which would give a tatol of 36. That einamitles 39 and 42. With 36, if we had 3 cells in each chamber, each side of the upper lveel would already have 9 of its 11 cells, meaning we’d have to leave empty cbhaemrs on the lower level. So 36 is also out. What about 33? We know we don’t want any sides with all 3′s, so we can place 2′s at opposite corners of the ueppr level.
Open Cloze
Up to this point we haven’t used the rule that each chamber must have 1 to 3 cells. It tells us that the _______ for the lower level is 8 cells, in which case, by rule 1, the upper level would have 16. On every side, the lower level would account for 3 of the 11 cells, so the upper would have to have 8. But if two opposite sides had 8, that would be the upper level’s entire 16, not leaving any for these two chambers. So 24 is out.
Let’s look at the other extreme. The upper level can have at most 3 times 8, or 24 cells, which would give a _____ of 36. That __________ 39 and 42. With 36, if we had 3 cells in each chamber, each side of the upper _____ would already have 9 of its 11 cells, meaning we’d have to leave empty ________ on the lower level. So 36 is also out. What about 33? We know we don’t want any sides with all 3′s, so we can place 2′s at opposite corners of the _____ level.
Solution
- upper
- eliminates
- chambers
- total
- minimum
- level
Original Text
Up to this point we haven’t used the rule that each chamber must have 1 to 3 cells. It tells us that the minimum for the lower level is 8 cells, in which case, by rule 1, the upper level would have 16. On every side, the lower level would account for 3 of the 11 cells, so the upper would have to have 8. But if two opposite sides had 8, that would be the upper level’s entire 16, not leaving any for these two chambers. So 24 is out.
Let’s look at the other extreme. The upper level can have at most 3 times 8, or 24 cells, which would give a total of 36. That eliminates 39 and 42. With 36, if we had 3 cells in each chamber, each side of the upper level would already have 9 of its 11 cells, meaning we’d have to leave empty chambers on the lower level. So 36 is also out. What about 33? We know we don’t want any sides with all 3′s, so we can place 2′s at opposite corners of the upper level.
Frequently Occurring Word Combinations
ngrams of length 2
collocation |
frequency |
upper level |
7 |
energy cells |
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Important Words
- account
- case
- cells
- chamber
- chambers
- corners
- eliminates
- empty
- entire
- extreme
- give
- leave
- leaving
- level
- meaning
- minimum
- place
- point
- rule
- side
- sides
- tells
- times
- total
- upper