The post Maths Week Scotland is back! appeared first on Making Maths Count.
]]>Maths Week Scotland will be back from 30 September – 6 October 2019.
You can uncover the maths we take for granted in our lives, and see maths in a whole new light during Maths Week Scotland. Even if you haven’t thought about maths since your last lesson at school, there is something for everyone.
Maths Week Scotland has a brand new website www.mathsweek.scot and twitter @MathsWeekScot where you can get involved. Schools can find resources, pin themselves on the map and get inspiration for their own Maths Week Scotland activities.
Events take place all across Scotland for families, adults and schools. Are you hosting a Maths Week Scotland event? Do you have an idea for a great event? Or want to see more in your area? Let us know at info@mathsweek.scot.
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]]>The post Deputy First Minister’s Holiday Maths Challenge – Summer 2019 solutions appeared first on Making Maths Count.
]]>Enjoy the rest of your summer!
DFM Summer Maths Challenge solutions
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]]>Please select this link to access the solutions
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]]>Christmas Challenge 2018 Solutions
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]]>The post Maths Week Scotland 2018 Solutions appeared first on Making Maths Count.
]]>Here are the solutions to all of this week’s puzzles.
(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, …
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.
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.
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.
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:
So the 6-layer stepped pyramid has 1+5+13+25+41+61=146 blocks.
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 |
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.
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]]>The post Maths Week Scotland – Poor Timekeeping appeared first on Making Maths Count.
]]>In cleaning out a drawer, Daniel found two old watches which he and his wife had discarded. He wound them up, and after setting them accurately, started both watches at the same time. An hour later he noticed that his old watch had gained a minute whilst his wife’s watch had lost two minutes. In fact his watch was running consistently fast and his wife’s watch was running consistently slow. Next morning, when he looked at the watches again, it was 7.20 on his old watch and 6.20 on his wife’s. What time was it when he started the watches running?
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]]>The post Maths Week Scotland – Coinage System appeared first on Making Maths Count.
]]>The imaginary country of Gaussland decides to adopt a new system of coinage. There are to be coins of ten different values (denominations). Those values (in cents) are:
The idea is that by changing to the new system it will be possible to pay for any item up to the value of 99 cents using no more than three coins. For example:
Show how no more than three coins can be used for each amount to 99 cents or, if this is not the case, find the smallest amount for which four coins are needed.
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]]>The post Maths Week Scotland – Stepped Pyramids appeared first on Making Maths Count.
]]>This stepped pyramid has been built by young children and their teacher using some of the building blocks in the classroom. (The blocks are all cubes of the same size.)
(a) How many blocks would be needed to build it?
(b) Now imagine that the children want to build a similar structure with twice as many layers. How many blocks would be needed?
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:
So the 6-layer stepped pyramid has 1+5+13+25+41+61=146 blocks.
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]]>The post Maths Week Scotland – Feeding the Horses appeared first on Making Maths Count.
]]>
During a very hard winter, Beth had only enough hay and corn to feed her six horses for another 30 days, and it would be another 75 days before spring would arrive. On the seventh day, before feeding time, she sold four of her horses.
Was she able to feed her remaining two horses for the rest of the winter?
Explain your reasoning.
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.
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]]>