Art of Problem Solving Series Prealgebra 2-Book Bundle Set (Textbook & Solutions Manual)

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Challenge Problems

On the next pour, i of the amount remaining is poured out, leaving % of the % container remaining.

Therefore, the container is % - g = % full after three pours.

We see a pattern! After the fourth pour, which is % of the remaining water, there is 31 - g = % of the water remaining. Similarly, after the fifth pour, % of the water remains. After the sixth pour, % of the water remains. After the seventh pour, % of the water remains. And after pour exactly % of the water remains.

15.38 Clearly all the numbers from 1 to 9 can be achieved, by pairing 1 with the appropriate number.

To count others in an organized manner, we make a table. We cross out all the single-digit numbers as well as any instance of a number after its first appearance:

x 2 3 4 5 6 7 8 9 2 4 6 8 1 0 1 2 1 4 1 6 1 8 9141513212427 3 4 162024283236 5 2530354045 36424854 6 7 495663 6472 3 9 31 We count 27 numbers in the table, and combining these with our 9 one-digit numbers, we have possible products.

15.39 We make an organized list of pictures. First, we consider the configurations in which the -

bottom left square is gray, since both of the restrictions on gray squares apply to this square.

Neither the square above it nor the square to its right can be white, so those two squares must also be gray. The upper right corner is both above and to the right of a gray square, so it must be gray, too. This gives us the one configuration shown on the right.

Next, we let the bottom left be white and the upper left be gray. The upper right then H

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must be gray, too, but there’s no restriction on the bottom right square. This gives us the two configurations shown on the right.

Having covered the cases in which at least one of the left squares is gray, we now consider those in which the two leftmost squares are white. We cannot have the bottom

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right square gray and the upper right square white, but the other three configurations with both squares on the left white are OK. This gives us a total of IE] possible configurations.

15.40 Working forwards from 1, it’s not at all clear what a good strategy is to get to 200. But if we work backwards, it becomes more clear. So, instead going from 1 to 200 by adding 1 or doubling at each step, we start from 200 and work our way down to 1 by subtracting 1 or halving at each step.

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