2018 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 20 Hard

A scanning code consists of a 7\times 7 grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of 49 squares. A scanning code is called symmetric if its look does not change when the entire square is rotated by a multiple of 90^\circ counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes? (2018 AMC 10A Problem, Question#20)

  • A.

    510

  • B.

    1022

  • C.

    8190

  • D.

    8192

  • E.

    65,534

Answer:B

Draw a 7\times 7 square.

\rm K\rm J\rm H\rm G\rm H\rm J\rm K
\rm J\rm F\rm E\rm D\rm E\rm F\rm J
\rm H\rm E\rm C\rm B\rm C\rm E\rm H
\rm G\rm D\rm B\rm A\rm B\rm D\rm G
\rm H\rm E\rm C\rm B\rm C\rm E\rm H
\rm J\rm F\rm E\rm D\rm E\rm F\rm J
\rm K\rm J\rm H\rm G\rm H\rm J\rm K

Start from the center and label all protruding cells symmetrically. (Note that "I" is left out of this labelling, so there are only 10 labels, not 11, as ending in K would suggest!)

More specifically, since there are 4 given lines of symmetry (2 diagonals, 1 vertical, 1 horizontal) and they split the plot into 8 equivalent sections, we can take just one-eighth and study it in particular. Each of these sections has 10 distinct sub-squares, whether partially or in full. So since each can be colored either white or black, we choose 2^{10}=1024 but then subtract the 2 cases where all are white or all are black. That leaves us with \boxed{\rm (B)~1022}.

There are only ten squares we get to actually choose, and two independent choices for each, for a total of 2^{10}=1024 codes. Two codes must be subtracted (due to the rule that there must be at least one square of each color) for an answer of \boxed{\rm (B)~1022}.

Imagine folding the scanning code along its lines of symmetry. There will be 10 regions which you have control over coloring. Since we must subtract off 2 cases for the all-black and all-white cases, the answer is 2^{10}-2=\boxed{\rm (B)~1022}.

Note that this problem is very similar to the 1996 AIME Problem 7.

This 7\times 7 square drawn in Solution 1 satisfies the conditions given in the problem. Calculating the number of ways of coloring it will solve the problem.

\rm K\rm J\rm H\rm G\rm H\rm J\rm K
\rm J\rm F\rm E\rm D\rm E\rm F\rm J
\rm H\rm E\rm C\rm B\rm C\rm E\rm H
\rm G\rm D\rm B\rm A\rm B\rm D\rm G
\rm H\rm E\rm C\rm B\rm C\rm E\rm H
\rm J\rm F\rm E\rm D\rm E\rm F\rm J
\rm K\rm J\rm H\rm G\rm H\rm J\rm K

In the grid, 10 letters are used: A, B, C, D, E, F, G, H, J, and K. Each of the letters must have its own color, either white or black. This means, for example, all K's must have the same color for the grid to be symmetrical.

So there are 2^{10} ways to color the grid, including a completely black grid and a completely white grid. Since the grid must contain at least one square with each color, the number of ways is 2^{10}-2=1024-2=\boxed{\rm (B)~1022}.

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