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Math Kangaroo practice

Name: ______________________Date: ____________3 questions · 15 points

1.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
2.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
3.How many diagonals does a convex polygon with 12 sides have? A diagonal joins two vertices that are not already joined by a side.[5]
- A54
- B66
- C108
- D120
- E42

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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 — 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.
2.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.
3.A — 54KA-0177Overcount, then correct
- BI counted every line joining two vertices and forgot to remove the 12 sides.
- CI counted each diagonal from both of its endpoints and forgot to halve.
- DI used 12 x 10 without halving, double counting every diagonal.
- EI subtracted 24 rather than 12, removing each side twice.

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