Kangaroo 2020 – Junior (secondary years 1 and 2)

Solution: Ten squares have a total perimeter of \(10\cdot4=40\) cm. The figure has 11 shared sides and each is counted twice: \(40-2\cdot11=18\) cm.

Solution: The values are 2346, 357, 168, 1239 and 12345. In order: 168, 357, 1239, 2346, 12345, so the middle one is \(1234+5=1239\).

Solution: The mother of Ann’s mother is her grandmother. The grandmother’s daughter is Ann’s mother or aunt, and that daughter’s mother is again the grandmother.

Solution: Each stripe runs from the left edge of the shirt, round the back, to the right edge. Now the right side is shifted by one stripe, so the end of each stripe joins the start of the next one. All stripes form one spiral with two free ends, so there is no closed ring.

Solution: \(\overline{AD}+\overline{CD}+\overline{AB}+\overline{CB}=20A+20C+2B+2D=2(\overline{AB}+\overline{CD})=2\cdot79=158\).

Solution: \(n+(n+1)+(n+2)+(n+3)=4n+6=2\Rightarrow n=-1\). The numbers are \(-1,0,1,2\).

Solution: The next such year is 2121, that is \(2121-2020=101\) years later.

Solution: Let her have \(k\) squares. After cutting there are \(k-3\) squares and \(10-k+6\) triangles: \(4(k-3)+3(16-k)=42\Rightarrow k+36=42\Rightarrow k=6\). She had \(10-6=4\) triangles.

Solution: A segment joining the midpoints of two sides of a triangle is half of the third side. So two pieces are 60 cm and two are 40 cm: \(120+80+2\cdot60+2\cdot40=400\) cm.

Solution: The two top points are 1 apart and the point on the right is 1 lower: \(A=\frac{1\cdot1}{2}=\frac12\). The other triangles have areas \(\frac32\), \(\frac92\) and \(\frac{11}2\).

Solution: 18 days are 2 full weeks (6 story days) and 4 more days. Four days in a row have the most stories from Saturday to Tuesday (3). If she arrives on Saturday she gets those 4 days first: 9 stories in all.

Solution: \(abcd=2cd\cdot cd=2(cd)^2\). \(50=2\cdot5^2\), \(200=2\cdot10^2\), \(450=2\cdot15^2\), \(800=2\cdot20^2\), but \(100=2\cdot50\) and 50 is not a square.

Solution: From the left signpost: Atown–Betown \(=3+1=4\) km. The right signpost is 6 km from Atown, so Betown is \(6-4=2\) km behind.

Solution: The other sides are \(x\) and \(\frac25x\), so one of them is 20. If \(x=20\): sides 20, 20, 8 and perimeter 48 cm. If \(\frac25x=20\): sides 20, 20, 50, which is not a triangle (\(20+20<50\)).

Solution: Adding the four diameters counts each number on the circle once and the centre 4 times: \(40+4c=4\cdot13=52\Rightarrow c=3\).

Solution: We need a two-digit \(n\) with \(20n=4\cdot5n\) a square, so \(5n\) is a square: \(n=5m^2\). These are \(n=20,45,80\): the numbers 2020, 2045 and 2080.

Solution: The left side of the triangle is inclined at \(60^\circ\) and the top side of the large square makes \(70^\circ\) with it, so it slopes \(10^\circ\) downwards. Then the right side of the square makes \(40^\circ\) with the right side of the triangle. The side of the small square is perpendicular to the side of the triangle: \(90^\circ-40^\circ=50^\circ\).

Solution: After 55 km he has \(14-5.5=8.5\) litres, enough for 85 more km, and 465 km remain. A full tank (40 litres) lasts 400 km, so the station must be at least \(465-400=65\) km and at most 85 km ahead: 75 km.

Solution: Divide by 17: \(x+3y=6\). Then \(9x+27y=9(x+3y)=54\).

Solution: The side is 9 dm and each triangle has area \(81:6=13.5\) dm². The top triangle has base 9 dm: \(\frac{9h}{2}=13.5\Rightarrow h=3\) dm to the top edge, so \(9-3=6\) dm to the bottom.

Solution: The digit sum is 45, so the number is always divisible by 9. It is divisible by 18 when it is even, that is when it ends in 2, 4, 6 or 8: \(\frac49\).

Solution: While the tortoise runs 5 km, the hare runs 25 km. If he ran \(x\) km sideways: \(x+\sqrt{x^2+25}=25\Rightarrow x^2+25=(25-x)^2\Rightarrow x=12\). The distance to the finish is \(25-12=13\) km.

Solution: From 1): a triangle is not large, so it is small. From 2): a blue shape is a triangle, so it is small. The other statements need not hold (for example there may be a small red triangle).

Solution: The overlap is a rhombus (both strips are 3 cm wide). If its side is \(a\): \((9-a)^2+3^2=a^2\Rightarrow90=18a\Rightarrow a=5\). The area is \(5\cdot3=15\) cm².

Solution: If the apex is \(v\), the four side faces give \(4v+2(15-v)=2v+30\) in total. If 7, 8, 9, 10 are the side faces: \(2v+30=34\Rightarrow v=2\) and the base sum is \(15-2=13\). The base cannot be one of them: then \(v=5\) and the smallest side sum is \(5+1+2=8\), so 7 is impossible.

Solution: Best is to paint two opposite faces and one between them: \(12+12+8=32\). If the three faces meet at a corner there are only \(3\cdot9=27\).

Solution: Let the sum be \(S\). The empty cell in the first row is \(S-10\), so the second column gives 1 in the last row. The third and fourth columns give \(S-15\) for their empty cells. Last row: \(x+1+7+(S-15)=S\Rightarrow x=7\).

Solution: There are \((10+15+17):2=21\) matches, so Ann rested 11 times. Nobody rests two matches in a row, so Ann rested in matches 1, 3, …, 21 and played in 2, 4, …, 20. She rested after each of her matches, so she lost them all, including the second.

Solution: The chord from \(A\) makes angle \(\alpha\) with the diameter, so the arc to the first vertex is \(180^\circ-2\alpha\). All upward segments are parallel, and so are all downward ones, so neighbouring vertices are always this arc apart. The four vertices cut the semicircle into 5 equal arcs: \(5(180^\circ-2\alpha)=180^\circ\Rightarrow\alpha=72^\circ\).

Solution: None ends in 0, so the last digits are 1–8 or 2–9. For 1–8: \(10k+d\) is divisible by \(d\) when \(d\mid10k\), and \(d=3,4,7,8\) need \(84\mid k\): \(k=84\), the numbers 841, …, 848. (For 2–9 we need \(252\mid k\), which gives four-digit numbers.) The digit sum of 841 is 13.