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Support The Journal Put your giant abacus away, the answers to Maths Week 2016 are here.
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# Maths Week: The solutions to your puzzles

Tens of thousands of you took our mathematical challenge all week. Here are the answers.
Oct 23rd 2016, 10:00 AM 16,615 8

THIS HAS BEEN Maths Week and if you’ve clicked in here, you are likely to have been puzzling over our fiendish maths challenges on TheJournal.ie all week.

So let’s not delay further: these are the answers you’ve been waiting for.

Let us know how you did in the comments…

Monday: The One With The Old Library Book Fines.

The answer: It would make a considerable dent in your inheritance. The fine due this week would be over €112 million – that is IR£88 million.

The method:

Compound interest is calculated as follows: A = P(1+I) n

• Where A = amount due
• P = principal (starting amount) £0.05
• I = interest rate (as decimal) 0.01
• n = number of intervals where interest is charged (2140 weeks)
• (41 years x 365 days + 11 leap days + 4 days to this Monday)/7 weeks
• A = 0.05(1.01) 2140 = IR£ 88,452,456

It will be obvious that adding interest after every week causes the amount to increase rapidly. A lot of people are caught out by seeming small interest rates but compounded over short intervals. It may be of little significance if paid off quickly but if left may accrue rapidly.

Tuesday: The One With The Poles Along A Road.

The method:

There are 10 poles and you run from 1 st to 10 th – that is 9 intervals which is 450m.

Speed is distance/time (we want the answer in km/h so it is convenient now to convert to km and hours)

450m is 0.45km. 3 minutes is 1/20 of an hour.

Speed is therefore is 0.45/(1/20) = 0.45 x 20 = 9 km/h

Wednesday: The One With The Four-Digit Number.

The method:

Assign the symbols w,x,y,z for the digits

• Number A is wxyz (this is not a multiplication – the symbols here represent the place digits)
• Number B is zwxy
• Number A , wxyz = 1000w +100x + 10y + z
• On rearranging into B it becomes: 1000z + 100w + 10x + y
• And we know that: (1000z + 100w + 10x + y) = 1.75(1000w +100x + 10y + z)
• (6000 + 100w + 10x + y) = 1.75(1000w +100x + 10y + 6)
• We also know that the second number is the first one reversed so therefore we can surmise: A = wxyz B = zwxy
• A reversed is zyxw
• Therefore zyxw = zwxy
• So y = w
• zyxw = 1.75 (wxyz)
• 6yxy = 1.75(yxy6)
• 6000 + 100y + 10x + y = 1.75(1000y + 100x + 10y + 6)
• 6000 + 101y + 10x = 1750y +175x + 17.5y + 10.5
• Multiply by 2 to get rid of unsightly decimals: 12000 + 202y + 20x = 3500y + 350x + 35y + 21
• This simplifies to: 11,979 = 3333y + 330 x
• There are infinite solution to this equation but we know that x and y must be integers.
• On inspection of the equation y dominates being a little more than a third of 11,979 – so must be 1,2, or 3. Starting with 3 will give 9999 for 3333y with remainder 1,980, so: 1,980 = 330x
• Which means x = 6.
• So starting number = 3636

Thursday: The One Where Two Cars Pass Through Toll Booths.

The method:

(It might be easier to think of two cars passing each other – how far apart would they be in 36 mins – this is an equivalent question)

They are driving towards each other at a rate of 115 + 110 = 225 km/h

Speed = distance/time

So distance = speed x time

36 mins = 36/60 hours =0.6 hours

Distance = 225 x 0.6 = 135 km

Friday’s Question: The One Where Two Kayakers Paddle On The Liffey.

The answer: Kayaker A. Kayaker B can only manage 6km/h in still water.

The method:

Kayaker A with an effort that would transport him at 9 km/h in still water against a current of 5 km/h can only manage 4 km/h.

Kayaker B is assisted by the current and will be travelling 5km/h faster than his normal ability.

We must calculate how far A gets in 10 minutes. (1/6 hour)

At 4km/h in 1/6 h, he will travel 4 x 1/6 = 4/6 km

Kayaker B will therefore have travelled a distance of (2.5 – 4/6) km in 10 minutes

Which will be: (2.5 – 4/6)/(1/6) = 11 km/h

If Kayaker B is doing 11 km/h assisted by the 5km/h current then his ability on still water is 6km/h.

Saturday morning: The One With The Tombola Tickets.

The answer: 63: 4 ending in 0, 14 ending in 5, 24 ending in 1 and 21 others.

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The method: Look for useful information in question and express it in equations –

• A = number of tickets ending in 0
• B = number of tickets ending in 5
• C = number of tickets ending in 1
• D = number of other tickets

Two out of three chances of winning gives us: 3(A+B+C) = 2(A + B + C + D) (1)

If you bought 60 tickets you would be sure of winning a bottle of brandy – therefore there are 59 tickets that are not brandy. B+C+D =59 (2)

For the others:

• A + C + D = 49 (3)
• A+B +D =39 (4)

• B+C+D =59 (2)
• A+ C +D =49 (3)
• A+B+D = 39 (4)

and we get: 2A + 2B + 2C + 3D = 147 (5)

But equation (1) tells us: 3(A+B+C) = 2(A + B + C + D) (1)

Multiplying out: 3A+3B+3C=2A+2B+2C+2D

Which reduces to: A+B+C =2D

Which when multiplied by 2 becomes: 2A + 2B + 2C = 4D (6)

Recalling that (5): 2A + 2B + 2C + 3D = 147 (5)

Substituting (6) into (5) gives us: 7D = 147

So that gives D – the number of non-winning tickets: D = 21

Substituting in this value into equations (2) (3) and (4):

• B+C = 38 (7)
• A+C = 28 (8)
• A+B = 18 (9)

Subtracting (8) from (7) gives: B-A=10 (10)

Adding (10) to (9) gives: 2B = 28

• Therefore B the number of tickets ending in 5 reveals itself as: B = 14
• Therefore we can find A, C: A = 4 and C = 24
• And total number of tickets A+B+C+D = 63

Saturday evening: The One With The 3-D Puzzles

Part one: B – Cannot be folded into a cube

Part two: Playing Die – A=3, B=6, C=2

### The maths gene? It all adds up to a myth>  