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Solutions for Session 2, Part D
See solutions for Problems: D1 | D2 | D3 | D4 | D5
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Problem D1 | |
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One method is that the number of blocks for the nth staircase is n more than the number of blocks for the (n - 1)st staircase, because you need to add on a row of n blocks to build the new staircase. This allows you to construct a table. One clever trick is to piece together two consecutive staircases; the 5th staircase fits nicely into the 6th staircase to form a 6-by-6 square!
<< back to Problem D1
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Problem D2 | |
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There will be one new row in staircase n + 1, a row with n + 1 blocks in it. That means that to go from staircase n to staircase n + 1, you will need to add n + 1 blocks.
<< back to Problem D2
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Problem D3 | |
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In the 274th staircase there will be 37,401 + 274 = 37,675 blocks. In the 275th staircase there will be 37,675 + 275 = 37,950 blocks. This answer can also be reached by using (275) * (276) / 2 = 37,950.
<< back to Problem D3
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Problem D4 | |
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There are 5,050 blocks in the 100th staircase. The fastest way is to use n(n + 1)/2.
<< back to Problem D4
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Problem D5 | |
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A way to build a general formula is to notice that two copies of the 6th staircase will form a 6-by-7 rectangle. For the nth staircase this will be n(n + 1) blocks for two staircases, which means n(n + 1) / 2 for one.
<< back to Problem D5
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