Parallelogram 42 • Level 4 • 19 Jun 2025Cheryl's birthday
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. Intermediate Maths Challenge Problem (UKMT)
1.1 April, May and June have 90 sweets between them.
May has three-quarters of the number of sweets that June has.
April has two-thirds of the number of sweets that May has.
How many sweets does June have?
- 60
- 52
- 48
- 40
- 36
- (Not answered)
Let
It follows that May has
Hence April has
So April has
Because April, May and June have 90 sweets between them:
Thus
and we deduce that
Hence June has 40 sweets.
2. Intermediate Maths Challenge Problem (UKMT)
2.1 How many positive cubes less than 5000 end in the digit 5?
- 1
- 2
- 3
- 4
- 5
- (Not answered)
The cube root of such numbers would have to end in a 5.
The cube of a positive integer ends in the digit 5 only when the positive integer itself ends in the digit 5.
So we check the cubes of the positive integers that have units digit 5, until we reach a cube which is greater than 5,000.
We have
We deduce that there are two positive cubes that end in the digit 5.
2.2 What proportion of all the cubes in the range from 1 to 1,000,000 end in the digit 7?
1 7 1 5 1 10 1 49 - None
- (Not answered)
For a cube number to end in 7, the original number must end in 3.
There are exactly 100 cubes from 1 to 1,000,000:
Only the final digit of the number that is cubed determines the final digit of the cubed number.
The only way for the cubed number to end in 7 is if the number that is cubed ends in 3.
This is true of one in every ten positive integers, and therefore true of a tenth of the integers from 1 to 100.
3. Cheryl’s birthday
Here is a birthday riddle that will put your mind in a spin.
The problem went viral several years ago because many people couldn’t quite believe the answer.
When you first see the problem, you too might think there just isn’t enough information, or you may get a different answer altogether.
This is where logic comes in - by thinking through each step very carefully, we can slowly but surely solve the mysterious case of Cheryl’s birthday.
Watch the video to see the problem, and the breakdown of the solution - are you convinced?
(If you have problems watching the video, right click to open it in a new window)3.1 What information was Bernard given by Cheryl?
- The month of her birthday
- The day of the month of her birthday
- The year of her birthday
- Her age
- (Not answered)
3.2 When Albert says he knows that Bernard doesn’t know Cheryl’s birthday, what information does this give to Bernard?
- Cheryl’s birthday is in May
- Cheryl’s birthday is in July
- Cheryl’s birthday is not in May or June
- It does not reveal any new information
- (Not answered)
Albert can only be sure that Bernard can’t know if the month (which Cheryl revealed to Albert) contains more than one date.
3.3 When Bernard then says he now knows the birthday, what information does this give to Albert?
- Cheryl’s birthday is not on the 14th
- Cheryl’s birthday is not on the 16th
- Cheryl’s birthday is in August
- It does not reveal any new information
- (Not answered)
3.4 How certain can Albert and Bernard be that they know Cheryl’s birthday?
- Around 90% certain
- 50%
- 100% - they have definitely worked it out
- Albert has a 25% chance of knowing the answer (since there are 4 months and he knows the month) and Bernard has a 16.7% chance (since there are 6 days and he knows the day)
- (Not answered)
4. Intermediate Maths Challenge Problem (UKMT)
4.1 The diagram shows an isosceles right-angled triangle which has a hypotenuse of length
The interior of the triangle is split up into identical squares and congruent isosceles right-angled triangles.
What is the total shaded area inside the triangle?
y 2 2 y 2 4 y 2 8 y 2 16 y 2 32 - (Not answered)
By Pythagoras, the other two sides of the triangle have length
Use this to first calculate the area of the triangle.
Let the two equal sides of the large right-angled triangle have length
By Pythagoras’ Theorem,a pplied to this triangle, we have
Therefore
It follows that the area of the large right-angled triangle is given by:
The shaded area is made up of four of the small squares.
The unshaded area is made up of two of the small squares and four half squares, which is the same as the area of four of the small squares.
Hence the shaded area is equal to the unshaded area.
Therefore the shaded area is half the area of the large right-angled triangle.
It follows that the shaded area is given by
5. Puzzle
5.1 There is a unique set of three different positive whole numbers whose sum is equal to their product.
What is the sum (and product)?
Correct Solution: 6
1 + 2 + 3 = 1 × 2 × 3 and this is the only possibility
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.