a) Assume that the pattern above continues. If a figure has 8 shaded hexagons, how many white hexagons are needed?

b) If a figure uses 32 white hexagons, how many shaded hexagons are in the figure?

c) Which figure has 29 white hexagons?

 

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1 Answer

Every time a black hexagon is added two white ones are also added, so we have the series:

1B+6W, 2B+8W, ..., nB+(6+2(n-1))W, where n is the number of blacks. In words this is: take 1 off the number of blacks and double it to get the additional whites. We started with 6 whites so we add 6 to this result.

So, all we need to find out how many whites there are is to use a linear equation: W=2(B-1)+6 or W=2B+4, where integer B>0. (We could have used x and y instead of B and W for black and white, but the initial letters may be more meaningful in this problem.)

(a) When B=8, W=20.

(b) When W=32, 32=2B+4, 2B=28, so B=14.

(c) No figure has 29 white hexagons because there are always an even number of whites: W=2B+4=2(B+2) which is an even number.

by Top Rated User (1.1m points)

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