Parallelogram 34 Year 8 2 Jul 2020Miracle Chocolate

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Noun: Parallelogram Pronunciation: /ˌparəˈlɛləɡram/

  1. a portmanteaux 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.

It’s the last normal Parallelogram of the school year... so it’s a toughie - and even better there is a super-giant Parallelogram next week to keep you busy over the summer holidays. Actually, it is not super-giant, just bigger than a normal Parallelogram. That’s because it is important that you do a bit of mathematics every week between the end of the summer term and the start of the autumn term to stop your brain from turning to sludge. Good luck!

1. Paradoxes

Last week, I showed you the missing square illusion, in which some shapes are re-arranged … and somehow the area at the end is less the area at the start. It seemed impossible.

Here is the same illusion made real with the help of a bar of chocolate.

These two videos (by James Grime) present the mystery in a different way and explains the maths behind the illusion.

The missing square illusion could be called a “paradox”, which means “a statement that contradicts itself or a situation which seems to defy logic.”

Here is an example of a paradox:

Your mission is not to accept the mission. Do you accept?

If you accept the mission, then you have failed, because the mission is to not accept the mission. But if you do not accept the mission, then you also have failed, because you did not accept the mission.

Try this one:

Answer truthfully (yes or no) to the following question: Will the next word you say be 'no'?

If your next word is “yes” then you have lied, because your next word was not “no”. And if your next word is “no”, then you have lied, because you said that your next word would not be “no”, but it was “no”.

And paradoxes can also come in the form of pictures. Take a look at this odd elephant.

3 marks

1.1 What is odd about the elephant?

  • Its tusks are uneven.
  • Its ears are too small.
  • It does not have a tail.
  • It has too many legs, or maybe not enough.
  • It has a very poor memory.

2. Junior Maths Challenge Problem (UKMT)

2 marks

2.1 The approximate cost of restoring the Flying Scotsman was £4 million. This was about 500 times the cost of building the steam engine in 1923.

Roughly what did the engine cost to build?

  • £800
  • £2000
  • £8000
  • £20,000
  • £80,000

There are 4,000,000 pounds in £4 million. Therefore, the cost of building the engine, in pounds, was roughly

1500×4000000=4000000500=8000.

So, roughly, the engine cost £8000 to build.

3. Junior Maths Challenge Problem (UKMT)

3 marks

3.1 How many pairs of digits (p,q) are there so that the five-digit integer ‘p869q’ is a multiple of 15?

  • 2
  • 3
  • 4
  • 5
  • 6
Show Hint (–1 mark)
–1 mark

If the number is a multiple of 15, then it must also be a multiple of both 3 and 5, so the last digit (q) must be 0 or 5.

Show Hint (–1 mark)
–1 mark

If q=0, then what values for p would make the number divisible by 3?

If q=5, then what values for p would make the number divisible by 3?

How many combinations of p and q do you have?

We have 15=3×5. Also, 3 and 5 have no common factors other than 1. It follows that the multiples of 15 are just the numbers that are multiples of both 3 and of 5.

For ‘p869q’ to be a multiple of 5 we require that its units digit q is either 0 or 5.

For ‘p869q’ to be a multiple of 3 we require that the sum of its digits is a multiple of 3.

We first consider the case where q is 0. In this case the sum of the digits of ‘p869q’ is given by

p+8+6+9+0=p+23.

Because ‘p869q’ is a five-digit integer, p is not 0. So the possible values of p are 1, 2, 3, 4, 5, 6, 7, 8 and 9. It is easily checked that there are three of these digits, namely 1, 4 and 7, for which p+23 is a multiple of 3.

Similarly, when q is 5, the sum of the digits of ‘p869q’ is

p+8+6+9+5=p+28.

In this case there are also three choices of p, namely 2, 5 and 8, for which p+28 is a multiple of 3.

It follows that the number of pairs of the digits (p,q) for which ‘p869q’ is a multiple of 15 is 3 + 3, that is, 6.

[Note: as we have found above, these pairs are (1, 0), (4, 0), (7, 0), (2, 5), (5, 5) and (8, 5).]

4. Junior Maths Challenge Problem (UKMT)

3 marks

4.1 The areas of the two rectangles in the diagram are 25 cm2 and 13 cm2 as indicated.

What is the value of x?

  • 3
  • 4
  • 5
  • 6
  • 7

The rectangle with area 13 cm2 has width 4 cm. Therefore the height of this rectangle is (13 ÷ 4) cm, that is, 134 cm.

It follows that the height of the rectangle with area 25 cm2 is 3+134 cm.

Because this rectangle has area 25 cm2, its width is 25÷3+134 cm.

Now 3+134=124+134=254. It follows that 25÷3+134=25÷254=25×425=4.

Hence the value of x is 4.

5. Junior Maths Challenge Problem (UKMT)

3 marks

5.1 Ali wants to fill the empty squares so that the number in each square after the fourth from the left is the sum of the numbers in the four squares to its left.

What number should Ali write in the final square?

  • 16
  • 8
  • 4
  • 2
  • 1
Show Hint (–1 mark)
–1 mark

Let the numbers that Ali needs to place in the empty squares be p, q, r and s, as shown in the figure.

Show Hint (–1 mark)
–1 mark

Because each of the numbers 8 and s is the sum of the four numbers that precede it,

0+q+r+1=8 (1)

and

q+1+r+8=s. (2)

Show Hint (–1 mark)
–1 mark

By subtracting equation (1) from equation (2), we have

q+1+r+80+q+r+1=s8,

Let the numbers that Ali needs to place in the empty squares be p, q, r and s, as shown in the figure.

Because each of the numbers 8 and s is the sum of the four numbers that precede it,

0+q+r+1=8 (1)

and

q+1+r+8=s. (2)

By subtracting equation (1) from equation (2), we have

q+1+r+80+q+r+1=s8,

and hence,

8=s8.

Therefore, by adding 8 to both sides, we deduce that

16=s.

It follows that Ali should write 16 in the final square.

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.