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Math Kangaroo practice

Name: ______________________Date: ____________5 questions · 20 points

1.What is the size of one interior angle of a regular twelve-sided polygon?[3]
- A150 degrees
- B30 degrees
- C165 degrees
- D1800 degrees
- E144 degrees
2.A sequence is defined by a(1) = 2 and a(n+1) = 2 a(n) - 1 for every n. What is a(6)?[3]
- A33
- B31
- C64
- D17
- E63
3.Each of the four edges of a square is painted either black or white. Two paintings count as the same if one can be rotated onto the other. How many genuinely different paintings are there?[4]
- A6
- B16
- C4
- D8
- E5
4.What is the smallest number n such that ANY collection of n whole numbers must contain two of them whose difference is divisible by 7?[5]
- A8
- B7
- C14
- D15
- E4
5.An 8 by 8 chessboard has two opposite corner squares removed, leaving 62 squares. Each domino covers exactly two squares that share an edge. Can the 62 squares be covered exactly by 31 dominoes?[5]
- ANo, because the two removed corners share a colour, so 32 squares of one colour remain and only 30 of the other
- BYes, because 62 is even and 31 dominoes cover exactly 62 squares
- CNo, because 62 is not divisible by 4
- DYes, but only if the dominoes may be placed diagonally
- ENo, because the board is no longer rectangular

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Answer key — Math Kangaroo practice

Each wrong choice carries the thinking that produces it. When a student picks C, this is what they were doing.

1.A — 150 degreesKA-0170Angles on lines, in triangles, in polygons
- BI found the exterior angle and answered with that instead of the interior one.
- CI used 360 divided by 12 subtracted from 195, or otherwise slipped a step in the arithmetic.
- DI gave the total of all twelve interior angles rather than one of them.
- EI used a ten-sided polygon by mistake.
2.A — 33KA-0171Recursive rules
- BI used the rule 2a(n) - 1 but started the sequence at 1 instead of 2.
- CI doubled six times and ignored the minus one entirely.
- DI stopped at a(5), one term early.
- EI computed 2^6 - 1, applying the minus one only once at the very end.
3.A — 6KA-0173Counting up to symmetry
- BI counted every colouring of the four edges and forgot that rotations make some of them identical.
- CI divided 16 by 4, but that only works when no painting is left unchanged by a rotation, and some are.
- DI halved 16, treating only the 180 degree turn as a symmetry.
- EI listed by how many edges are black but forgot that two black edges can be adjacent or opposite.
4.A — 8KA-0175The pigeonhole principle
- BI used the number of possible remainders without adding one for the pair that must collide.
- CI doubled 7, thinking I needed two full sets of remainders.
- DI doubled 7 and added one, applying the pigeonhole idea to the wrong number of boxes.
- EI guessed from small cases without identifying what the boxes actually are.
5.A — No, because the two removed corners share a colour, so 32 squares of one colour remain and only 30 of the otherKA-0176Coloring arguments
- BI checked that the counts match but a matching count does not make a covering possible.
- CI invented a divisibility condition; dominoes cover two squares, so only divisibility by 2 could matter.
- DI changed the rules rather than testing them; the question says dominoes cover squares sharing an edge.
- EI appealed to the shape, but plenty of non-rectangular regions can be tiled by dominoes.

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