Parallelogram 29 • Level 3 • 21 Mar 2024Paper and ink

Noun: Parallelogram Pronunciation: /ˌparəˈlɛləɡram/

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.

5 marks

1.1 How many of the six faces of a die have fewer than three lines of symmetry?

2
3
4
5
6
(Not answered)

Each of faces 1, 4 and 5 has four axes of symmetry, whilst each of faces 2, 3 and 6 has two axes of symmetry only.

2.

5 marks

2.1 A small ink cartridge has enough ink to print 600 pages. Three small cartridges can print as many pages as two medium cartridges. Three medium cartridges can print as many pages as two large cartridges. How many pages can be printed using a large cartridge?

1200
1350
1800
2400
5400
(Not answered)

Three small cartridges have enough ink for 3×600=1800 pages. So 1800 is the number of pages that two medium cartridges can print. Hence one medium cartridge can print 12 (1800) = 900 pages. So three medium cartridges have enough ink for 3×900=2700 pages. So 2700 is the number of pages that two large cartridges can print. Hence one large cartridge can print 12 (2700) = 1350 pages.

An alternative, algebraic method, is to let m be the number of pages that a medium cartridge can print, and l be the number of pages that a large cartridge can print. From the information we are given we have the equations 3×600=2m and 3m=2l.

From these we can deduce that l = 32 m = 32 × 3×6002 =9×150=1350.

3.

5 marks

3.1 The six-member squad for the Ladybirds five-a-side team consists of a 2-spot ladybird, a 10-spot, a 14-spot, an 18-spot, a 24-spot and a pine ladybird (on the bench). The average number of spots for members of the squad is 12. How many spots does the pine ladybird have?

4
5
6
7
8
(Not answered)

The total number of spots which the six ladybirds have is 6×12=72. So the number of spots which the pine ladybird has is 72−2+10+14+18+24=4.

4.

6 marks

4.1 The diagram is a “map” of Jo’s local rail network, where the dots represent stations and the lines are routes.

Jo wants to visit all the stations, travelling only by train, starting at any station and ending at any station, with no restrictions on which routes are taken.

What is the smallest number of stations that Jo must go to more than once?

1
2
3
4
5
(Not answered)

We label the stations as shown in the diagram on the right so that we can refer to them. Jo can visit the stations in the order:

O,P,Q,R,S,R,T,X,Y,Z,Y,X,W,V,U.

In this way she visits all the stations and goes through just three of them, R, Y and X more than once.

It remains only to show that Jo cannot visit all the stations without going through at least three of them more than once.

Jo must go through the junction stations R and X more than once. If Jo does not start or finish at O, she will have to go through P more than once. If she does not start or finish at U she will have to go through V more than once. If she starts at O and finishes at U, or vice versa, she will have to go through Y more than once. It follows that Jo has to go through at least three stations more than once. So this is the smallest number of stations Jo must go through more than once.

5.

6 marks

5.1 Peter wrote a list of all the numbers that could be produced by changing one digit of the number 200. How many of the numbers in Peter's list are prime?

0
1
2
3
4
(Not answered)

If the hundreds or tens digit of 200 is changed, but the units digit is unchanged, the resulting number is, like 200, a multiple of 10, and so cannot be prime. So we need only consider the 9 numbers that we can get by changing the units digit. Of these, we can see immediately that 202, 204, 206 and 208 are all divisible by 2 and 205 is divisible by 5. So none of them is prime.

This just leaves 201, 203, 207 and 209. Now:
201=3×67,
203=7×29,
207=3×3×23 and
209=11×19.

None of them are prime, so none of the numbers in Peter's list are prime.

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.