1994 AIME Problems/Problem 11
Problem
Ninety-four bricks, each measuring are to stacked one on top of another to form a tower 94 bricks tall. Each brick can be oriented so it contribues or or to the total height of the tower. How many different tower heights can be achieved using all ninety-four of the bricks?
Solution
We have the smallest stack, which has a height of inches. Now when we change the height of one of the bricks, we either add inches, inches, or inches to the height. Now all we need to do is to find the different change values we can get from 's, 's, and 's. Because , , and are all multiples of , the change will always be a multiple of , so we just need to find the number of changes we can get from 's, 's, and 's.
From here, we count what we can get:
It seems we can get every integer greater or equal to four; we can easily deduce this by considering parity or using the Chicken McNugget Theorem, which says that the greatest number that cannot be expressed in the form of for being positive integers is .
But we also have a maximum change (), so that will have to stop somewhere. To find the gaps, we can work backwards as well. From the maximum change, we can subtract either 's, 's, or 's. The maximum we can't get is , so the numbers and below, except and , work. Now there might be ones that we haven't counted yet, so we check all numbers between and . obviously doesn't work, does since 6 is a multiple of 3, does because it is a multiple of (and ), doesn't since is not divisible by or , does since , and and don't, and does.
Thus the numbers , , all the way to , , , , and work. That's numbers. That's the number of changes you can make to a stack of bricks with dimensions , including not changing it at all.
See also
1994 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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