Making Maths Count
Maths Week Scotland 2018 Solutions
September 16, 2018 by John Swinney MSP No Comments
I hope you’ve enjoyed this weeks Maths Week Scotland puzzles. How did you get on? I think they’ve been a lot of fun.
Here are the solutions to all of this week’s puzzles.
Solution to: Stepping Up
(a) 6
(b) 12
(c) No solution because Ross steps on odd-numbered steps with his left foot and Elena steps on only even-numbered steps with her left foot.
Ross puts his left foot on steps 1, 3, 5, … and his right foot on steps 2, 4, 6, …
Elena puts her left foot on steps 2, 6, 10, … and her right foot on steps 4, 8, 12, …
Tom puts his left foot on steps 3, 9, 15, … and his right foot on steps 6, 12, 18, …
Solution to: Dotty Rectangles
Answers: (a) 24 (b) 0, 11, 14, 15 (c) 136
(a) A 9×4 rectangle would have 24 dots inside (arranged in an 8 by 3 pattern).
(b) The dimensions of the rectangle could be 24×1,12×2,8×3 or 6×4. So the possible numbers of dots inside are 0, 11, 14, 15.
(c) If the perimeter is 54, then length + width = 27. Sharing 27 so that one number is twice the other gives 18 and 9. So the length is 18 and the width is 9. Therefore, the number of dots inside is 17×8=136.
Solution to: New Outfit
Answer: £145
Here are two possible ways to reach the answer:
1.
The trousers and shoes cost £115.
The shirt and the shoes cost £100.
So the trousers cost £15 more than the shirt.
But the trousers and the shirt cost £75. So we need two numbers that add to 75 but which differ by 15. They are 45 and 30.
So the shirt costs £30 and the cost of all three items is £115 + £30 = £145.
2.
Imagine that Asif and his friend, let’s call him Alistair, go together and buy a shirt, a pair of trousers and a pair of shoes each. As strange as it seems, their tastes are so similar that they select exactly the same items. Together they buy:
shirt + trousers | £75 |
shirt + shoes | £100 |
trousers + shoes | £115 |
£290 |
They have paid £290 between them, which is £145 each.
Solution to: Feeding the Horses
Answer: Yes
When Beth sold the horses, 69 days of winter remained and she had enough hay and corn to feed 6 horses for another 24 days. She could feed the remaining 2 horses for 24×3=72 days. So there was sufficient food.
Alternatively, at the start, Beth had 6×30=180 portions of horse food.
She used 6×6=36 portions whilst she still has 6 horses, so had 144 portions left.
With 69 days to go and 2 horses to feed, she needed 138 portions. So the food proved sufficient.
Solution to: Stepped Pyramid
Answers: 19 blocks, 146 blocks
(a) The pink layer on top has a single block.
The yellow layer has 5 blocks (one of which is hidden under the red block).
The blue layer has 1+3+5+3+1=13 blocks.
So the total number of blocks in the 3 layers of the stepped pyramid is 1+5+13=19.
(b) Layers 4, 5 and 6 (working downwards) will have the following numbers of blocks in them:
- 1+3+5+7+5+3+1= 25
- 1+3+5+7+9+7+5+3+1= 41
- 1+3+5+7+9+11+9+7+5+3+1= 61
So the 6-layer stepped pyramid has 1+5+13+25+41+61=146 blocks.
Solution to: Coinage System
Notes:
(1) Though the arithmetic required to solve this problem is straightforward, the amount of work involved is considerable. One way to handle this issue is to divide the task so that many individuals or groups work on a subset and then amalgamate the results. If this is still thought too onerous, there is a similar problem which uses the coins to 45 cents and asks for the first amount for which four coins are needed (to which the answer is 68 cents).
(2) The title of the problem invokes the name ‘Gauss’. The German mathematician, Carl Friedrich Gauss (1777–1855), is considered one of the greatest in the development of the subject. One of his results, discovered and proved when he was a teenager, was that every positive whole number can be written as the sum of at most three triangular numbers, which start:
The result is the basis for this question about coin denominations.
Answer: 95
All but two of the amounts to 99 cents are possible. The smallest amount for which four coins are needed is 95 cents and there are six different ways in which this can be done.
1 | 1 | 11 | 10,1 | 21 | 21 | 31 | 28,3 | 41 | 28,10,3 |
2 | 1,1 | 12 | 10,1,1 | 22 | 21,1 | 32 | 28,3,1 | 42 | 21,21 |
3 | 3 | 13 | 10,3 | 23 | 21,1,1 | 33 | 21,6,6 | 43 | 21,21,1 |
4 | 3,1 | 14 | 10,3,1 | 24 | 21,3 | 34 | 28,6 | 44 | 28,10,6 |
5 | 3,1,1 | 15 | 15 | 25 | 21,3,1 | 35 | 28,6,1 | 45 | 45 |
6 | 6 | 16 | 15,1 | 26 | 15,10,1 | 36 | 36 | 46 | 45,1 |
7 | 6,1 | 17 | 15,1,1 | 27 | 15,6,6 | 37 | 36,1 | 47 | 45,1,1 |
8 | 6,1,1 | 18 | 15,3 | 28 | 28 | 38 | 36,1,1 | 48 | 45,3 |
9 | 6,3 | 19 | 10,6,3 | 29 | 28,1 | 39 | 36,3 | 49 | 28,21 |
10 | 10 | 20 | 10,10 | 30 | 15,15 | 40 | 36,3,1 | 50 | 28,21,1 |
51 | 45,6 | 61 | 45,15,1 | 71 | 55,10,6 | 81 | 45,21,15 | 91 | 55,36 |
52 | 45,6,1 | 62 | 55,6,1 | 72 | 36,36 | 82 | 55,21,6 | 92 | 55,36,1 |
53 | 28,15,10 | 63 | 21,21,21 | 73 | 45,28 | 83 | 55,28 | 93 | 45,45,3 |
54 | 45,6,3 | 64 | 36,28 | 74 | 45,28,1 | 84 | 28,28,28 | 94 | 55,36,3 |
55 | 55 | 65 | 36,28,1 | 75 | 36,36,3 | 85 | 55,15,15 | 95 | |
56 | 55,1 | 66 | 45,21 | 76 | 55,21 | 86 | 55,21,10 | 96 | 45,45,6 |
57 | 55,1,1 | 67 | 45,21,1 | 77 | 55,21, 1 | 87 | 45,36,6 | 97 | 55,36,6 |
58 | 55,3 | 68 | 55,10,3 | 78 | 36,36,6 | 88 | 45,28,15 | 98 | 55,28,15 |
59 | 55,3,1 | 69 | 45,21,3 | 79 | 55,21,3 | 89 | 55,28,6 | 99 | |
60 | 45,15 | 70 | 55,15 | 80 | 55,15,10 | 90 | 45,45 |
Solution to: Poor Timekeeping
Answer: 11.00 a.m.
Every hour, Daniel’s watch gained 1 minute and his wife’s lost 2 minutes. So an hour later they were 3 minutes apart. If in the morning they are 1 hour apart, 20 hours must have elapsed. And the time when Daniel checked his old watch was not 7.20 but actually 7.00 because it had gained 20 minutes overnight.
Daniel started the old watches going 20 hours earlier, which was 11 a.m. the previous day.
Tags: John Swinney, Maths Week, puzzles
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