Maths Wizard Challenge

Here are the answers to this week’s tricky mathematical conundrums:

1) A rectangle has a length 4 more than its width, and a perimeter of 28. What is its area?

Answer: 45

Solution: The dimensions are 9 and 5, so the area is 45

2) A rectangle has a perimeter of 21 and an area of 5. What is its length?

Answer: 10

Solution: The dimensions are ½ and 10, so its length is 10

Well done if you worked out the right answers! Come back next Monday for another pair of questions to get you thinking!

1)  Two numbers have a sum of -2 and a product of -1.25. What is their difference?

Answer: 3

Solution: The two numbers are 0.5 and -2.5, and their difference is 0.5-(-2.5)=3

2) What is the smallest positive integer divisible by each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10?

Answer: 2520

Solution: 2x2x2x3x3x5x7=2520

3) How many different combinations of coins add to 7 pence?

Answer: 6

Solution: 7=5+2=5+1+1=2+2+2+1=2+2+1+1+1=2+1+1+1+1+1=1+1+1+1+1+1+1. There are six combinations overall.

4) In a shop, the price of a jumper is increased by a third. A week later, the price falls by a third. What fraction of the original price of the jumper is the current price?

Answer: 8/9

Solutions: After the price increase, the jumper’s price is 4/3 of the original price. This is equivalent to 12/9. So after the fall in price, the new price is 8/9 of the original price.

5)  In how many ways is it possible to pick three whole number angles to make an isosceles triangle?

Answer:  89

Solution: The base angles are equal, and can take any whole number value from 1 degree to 89 degrees, so this can be done in 89

6)  I have a circle made of paper. What is the smallest number of folds required to turn the circle into a polygon (a shape with straight edges)?

Answer: 3

Solutions:  Every edge requires a different fold, and every polygon has at least three sides, so the best we can hope for is three. And it is in fact possible to create a triangle with three folds; so the answer is three.

7)  Which negative number is four times as far away from 41 as from 2?

Answer:  -11

Solution: -11 (it differs from 2 by 13 and from 41 by 52)

8) What is the smallest cube divisible by 6 and 10?

Answer:  27000

Solution: The cube must be divisible by 2, 3 and 5. Further, in a cube, the exact power to which any prime appears in its prime factorisation must be a multiple of 3. So the smallest cube divisible by 6 and 10 is 2 cubed x 3 cubed x 5 cubed = 27000

9) Find three positive integers, each of which has an equal number of even factors and odd factors

Answer: Any number divisible by 2, except 4

Solution: Any number divisible by 2 but not by 4 will do. The first few examples are 2 (factors are 1,2), 6 (factors are 1, 2, 3, 6) and 10 (factors are 1, 2, 5, 10)

10) Find three positive integers, each of which has three times as many even factors as odd factors

Answer: Any number divisible by 8 but not by 16

Solution: Any number divisible by 8 but not by 16 will do. The first few examples are 8 (factors are 1, 2, 4, 8), 24 (factors are 1, 2, 3, 4, 6, 8, 12, 24) and 40 (factors are 1, 2, 4, 5, 8, 10, 20, 40)

11)  In how many ways can the letters in the word PUPIL be arranged?

Answer: 60

Solution: If all five letters were different, there would be 5x4x3x2x1=120 arrangements. But the two Ps are identical, so 120 double-counts the arrangements. The required total is 60

12)  In how many of these arrangements are the two Ps next to each other?

Answer: 24

Solution: Treating the Ps as a single object, there are 4x3x2x1=24 arrangements in which the Ps are next to each other

13) Six of the interior angles in a heptagon are 110, 120, 130, 140, 150 and 160 degrees. What is the size of the seventh angle?

Answer: 90

Solution: The interior angle sum in a heptagon is 5×180=900 degrees, and the six angles given add to 810 degrees, so the seventh angle measures 90 degrees

14) A quadrilateral with four right angles has an area of 12.25 square centimetres. What is its smallest possible perimeter?

Answer: 14cm

Solution: The smallest perimeter will occur when the quadrilateral is a square. As 12.25=49/4, the side of the square would be 7/2 cm, and the perimeter would be 4×7/2=14 centimetres.

Questions 19 & 20

Four football teams play each other exactly once, with 3 points for a win, 1 for a draw and 0 for a loss. The teams are then ranked according to the number of points. Teams with the same number of points are awarded the same rank.

15) What is the smallest possible total number of points for the highest ranked team(s)?

16) For the football teams in question 19, what is the greatest possible total number of points for the lowest ranked team(s)?

Solutions & Answers:

There are 6 games in total, and in each game, either 2 or 3 points are awarded overall. So the total number of points awarded is between 12 and 18, and the average number of points per team is between 3 and 4.5

15) The highest ranked team cannot have fewer points than the average number of points per team, so they must score at least 3 points. This can happen is every match is a draw. So the smallest possible number of points for the highest ranked team is 3

16) The lowest ranked team cannot have more points than the average number of points per team, so they cannot score more than 4 points. This can happen is every team wins one, draws one and loses one of their matches. So the greatest possible number of points for the lowest ranked team is 4

17) In the sequence 1, 2, 4, 5, 7, 8, …, after the first term we alternately add 1 to the previous term, then add 2 to the previous term. Does 999,999 appear in the sequence?

Answer: No

Solution: The sequence contains precisely the numbers that are not divisible by 3; but 999,999 is a multiple of 3. So 999,999 does not appear in the sequence

18) Today’s date, in the DD/MM/YYYY format, is 28/06/2021. Using the same format, what was the last date when every digit was different?

Answer: 25/06/1987

Solution: Nothing this century works, because the month must contain a 0, a 2, or two 1s. The most recent valid date is 25/06/1987

19) How many positive integers under 100 are divisible by 4 and 6?

Answer: 8

Solution: The numbers divisible by 4 and 6 are precisely the multiples of 12, of which there are eight under 100.

20) How many positive integers under 100 are divisible by 4 or 6?

Answer: 32

Solution: There are 24 multiples of 4 under 100, and 16 multiples of 6 under 100. Adding 24 and 16 gives a total of 40; but this includes every multiple 12 twice, of which there are eight. So the final answer is 24+16-8=32

21) How many positive integers under 100 are divisible by 4 and 6?

Answer: 8

Solution: The numbers divisible by 4 and 6 are precisely the multiples of 12, of which there are eight under 100.

22) How many positive integers under 100 are divisible by 4 or 6?

Answer: 32

Solution: There are 24 multiples of 4 under 100, and 16 multiples of 6 under 100. Adding 24 and 16 gives a total of 40; but this includes every multiple 12 twice, of which there are eight. So the final answer is 24+16-8=32

If you’d like to try some more tricky problem solving puzzles, here are some web sites you can look at. I’m certain you will find plenty to challenge you!

Click on any of the links to take you straight there.

Junior Maths Challenge

Here you can try online challenges from previous Junior Maths Challenges, as well as download past papers and solutions.

https://www.ukmt.org.uk/competitions/solo/junior-mathematical-challenge/archive

Nrich Website / Twitter

The NRICH Project aims to enrich the mathematical experiences of all learners. They have lots of activities, questions and games to develop your mathematical skills, whatever level you are. They definitely make maths fun!

https://nrich.maths.org/primary

They also post questions on twitter for all ages of students.

https://twitter.com/nrichmaths

Times Tables Rockstars

Times Tables Rock Stars is a carefully sequenced programme of fun daily times tables practice, which concentrates on a different times table each week. It has a small subscription of £7.20 a year for a family.

https://ttrockstars.com/home

As well as these web sites, I can also recommend the brilliant maths book Elastic Numbers, written by one of Hampton’s own Maths Teachers, Mr Griller.

Elastic Numbers is full of fun and challenging mathematical treats for the serious problem solvers among you! It’s available from most book shops and also online.