Parallelogram 20 Year 7 25 Feb 2021Optimising your pizza money

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Noun: Parallelogram Pronunciation: /ˌparəˈlɛləɡram/

  1. a portmanteau word combining parallel and telegram. A message sent each week by the Parallel Project to bright young mathematicians.
  • Tackle each Parallelogram in one go. Don’t get distracted.
  • Finish by midnight on Sunday if your whole class is doing parallelograms.
  • Your score & answer sheet will appear immediately after you hit SUBMIT.
  • Don’t worry if you score less than 50%, because it means you will learn something new when you check the solutions.

1. Crazy Countdown puzzle

This is a Countdown numbers game, that John solves in a ridiculously complicated way. The challenge is to make 813 from [1, 10, 25, 50, 75, 100]. Have a go at solving it before watching the video.

(If you have problems watching the video, right click to open it in a new window)

There is a simpler way of writing this solution that avoids heading towards such astronomical numbers – the six starting numbers are in bold:

First, 75 - 10 = 65
Then, 65 x 25 = 1625
Then, 1625 + 1 = 1626
Then, 100 / 50 = 2
Finally, 1626 / 2 = 813

3 marks

1.1 What is the biggest number that you can make with [1, 10, 25, 50, 75, 100] by adding, subtracting, dividing and multiplying? By the way, the answer is not obvious.

Correct Solution: 103,125,000

You might multiply all the numbers, which is 93,750,000, but then you are not making any use of the 1, because anything multiplied by 1 remains unchanged. You could multiply all the numbers and then add the 1 at the end, which is 93,750,001, but that does not increase the total much. Instead, add 1 to 10 first, and multiply everything, so you have 11 x 25 x 50 x 75 x 100 = 103,125,000.

If you add 1 to 10, the new number (11) is 10% bigger, so it makes the final total 10% bigger. If instead you add 1 to 100, the new number (101) is only 1% bigger, so its makes the final total only 1% bigger, which is 94,687,500.

2. Junior Maths Challenge Problem (UKMT)

2 marks

2.1 Gill leaves Lille by train at 09:00. The train travels the first 27 km at 96 km/h. It then stops at Lens for 3 minutes before travelling the final 29 km to Lillers at 96 km/h. At what time does Gill arrive at Lillers?

  • 09:35
  • 09:38
  • 09:40
  • 09:41
  • 09:43
  • (Not answered)

In total, the train travels 27 km + 29 km = 56 km.

So the combined time for these two parts of the journey is 5696 hours = 712 hours = 35 minutes.

There is also the 3 minutes of waiting at Lens. The total journey time, therefore, is 38 minutes. So Gill arrives at 09:38.

3. The pizza equation

Your teachers and parents might have told you about the many applications of mathematics, everything from engineering to computer science, but have they told you about pizza economics? This video explains the mathematics behind buying a pizza.

(If you have problems watching the video, right click to open it in a new window)

1 mark

3.1 Small pizzas are 30 cm in diameter, while super-sized pizzas are 90 cm in diameter. How many small pizzas do you need in order to get the same amount of pizza as one super-sized pizza?

  • 1
  • 3
  • 5
  • 7
  • 9
  • (Not answered)

The video explained that if the radius (or diameter) increases by a factor x, then the area increases by a factor x2. In this case the diameter increases by 3, so the area increases by 32 = 9.

2 marks

3.2 Medium pizzas are 45 cm in diameter and cost £10. If the pricing was completely fair (ie, cost per cm2 was constant), then how much should the small 30 cm diameter pizza cost?

  • £3.00
  • £4.44
  • £4.50
  • £5.55
  • £6.66
  • (Not answered)

The small pizza is 3045 or 23 as wide as the medium pizza, so its area is (23)2 smaller than the area of the medium pizza. Therefore, the small pizza SHOULD cost (23)2 × £10 = £4.44. In practice, it will probably cost more, because of the reasons given in the video.

4. Junior Maths Challenge Problem (UKMT)

3 marks

4.1 Evariste and Sophie both bought some stamps for their collections. Each stamp Evariste bought cost him £1.10, whilst Sophie paid 70p for each of her stamps. Between them they spend exactly £10.00. How many stamps did they buy in total?

  • 9
  • 10
  • 11
  • 12
  • 13
  • (Not answered)

Suppose that Evariste buys x stamps and Sophie buys y stamps. Then 1.10x + 0.70y = 10.00. Hence, 11x + 7y = 100. We have one equation with two unknowns, so we need to use the fact that in this problem x and y are non-negative integers. x and y are whole numbers and bigger than or equal to zero.

By re-arranging the equation, y=10011x7. We need to test different values of x and find a whole number value for x that gives a whole number value for y. In other words, we need to find a value for x, such that 10011x is divisible by 7 without a remainder.

Taking x=0,1,2,9, we have y=10011x=100,89,78,67,56,45,34,23,12,1. We see that only for x=4 is 10011x divisible by 7. So x=4 and y=567=8. So Sophie buys 8 stamps and Evariste buys 4 stamps, and they buy 12 stamps in total.

5. The monoalphabetic substitution cipher – how many keys?

Last week we looked at the monoalphabetic substitution cipher, which swapped one letter for a different letter. The exact way that letters are substituted is known as the key. If the sender uses a key to substitute one letter for a different letter, then the receiver uses the same key to reverse the substitutions and read the original message. Here is one possible key:

a b c d e f g h i j k l m
n o p q r s t u v w x y z

The challenge for an enemy agent who has intercepted a coded message is to work out the key and read the message. If you're the enemy agent with a coded message, how long would it take you to work out the key? How long would it take to work out how each letter has been substituted?

To answer that question, first we need to know how many different keys are possible. Another way to phrase this question is to ask: "How many different ways can you re-arrange an alphabet of 26 letters?"

This is a big question, so let’s start with a smaller alphabet and see if we can find a pattern.

  • A 1-letter alphabet (A) can only be arranged in 1 way – A.
  • A 2-letter alphabet (AB) can be arranged in 2 ways – AB, BA
  • A 3-letter alphabet (ABC) can be arranged in 6 ways – ABC, ACB, BAC, BCA, CAB, CBA.

A pattern is emerging, and you can see it below:

Size of Alphabet Number of arrangements Why? Factorial
1 1 1 1!
2 2 1 × 2 2!
3 6 1 × 2 × 3 3!

(1 × 2 × 3) is called “three factorial” and is abbreviated to 3!

But why is 3! the number of ways that you can arrange a 3-letter alphabet?

  • imagine an arrangement.
  • the first letter could be any of the 3 letters.
  • the second letter could be either of the other 2 letters.
  • the third letter must be the 1 remaining letter.
  • so we had 3 × 2 × 1 possibilities.
1 mark

5.1. How many ways can you arrange a 4-letter alphabet (ABCD)?

Correct Solution: 24

There are 4! ways to arrange a 4-letter alphabet, which is 24 or (4 × 3 × 2 × 1) possibilities.

1 mark

5.2. How many ways can you arrange a 5-letter alphabet (ABCDE)?

Correct Solution: 120

There are 5! ways to arrange a 5-letter alphabet, which is 120 or (5 × 4 × 3 × 2 × 1) possibilities.

2 marks

5.3. What is the value of 8! (or how many ways can you arrange an 8-letter alphabet?)

You can work it out using your calculator’s factorial button – look for the exclamation mark – or you can work it out by multiplying over and over again.

Correct Solution: 40320

2 marks

5.4. By now, you should have grasped that there are n! ways to rearrange n letters or n objects. Now think about how you would write down n! if you were breaking it down into a series of multiplications. How would you write it?

  • n! = n × (n – 1) × … × (n – n)
  • n! = (n – 1) × (n – 2) × … × 2 × 1
  • n! = n × (n – 1) × … × 2 × 1
  • (Not answered)

6. Back to our original question

1 mark

6.1. How many ways can you arrange our usual 26-letter alphabet? In other words, how many different keys are there for the monoalphabetic substitution cipher?

  • 26
  • 2626
  • 26 + 25 + ... + 2 + 1
  • 26!
  • 262
  • (Not answered)
2 marks

6.2. Convert your answer from question 6.1 into an actual number by using a calculator. The answer is more than 100 million billion billion - how many million billion billions is it? Please provide a three digit answer.

Correct Solution: 403 million billion billion

If you type 26! into a calculator, then the answer is 403,291,461,126,605,635,584,000,000 000 (or it will look something like 4.0329...e26), so the answer is 403 (million billion billion).

7. Negative numbers

2 marks

7.1 What is the value of the missing number? 1×2×3×4×__=120?

Correct Solution: 5

3 marks

7.2 What is the value of 110?

Correct Solution: 1

2 marks

7.3 What is the value of 3×15? Give your answer as a decimal.

Correct Solution: -0.6

2 marks

7.4 What is the value of 0.5×5×2?

Correct Solution: 5

2 marks

7.5 What is the value of 12342?

Correct Solution: 5

Before you hit the SUBMIT button, here are some quick reminders:

  • You will receive your score immediately, and collect your reward points.
  • You might earn a new badge... if not, then maybe next week.
  • Make sure you go through the solution sheet – it is massively important.
  • A score of less than 50% is ok – it means you can learn lots from your mistakes.
  • The next Parallelogram is next week, at 3pm on Thursday.
  • Finally, if you missed any earlier Parallelograms, make sure you go back and complete them. You can still earn reward points and badges by completing missed Parallelograms.

Cheerio, Simon.

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