Maths Wizard Challenge

This Week’s Maths Wizard Problem…

Hello young maths wizards!

Are you enjoying my mathematical challenges each week? If you are, I have 2 more here for you, so put on your mathematical thinking caps and have a go.

You can find the answers under the the Maths Wizard Answer tab – but remember, no peeking at the solution until you’ve tried your best!

If you particularly like solving mathematical problems or if you would just like a bit more practice, I have included some excellent maths websites with fun and challenging questions and games under the Further Resources tab.

So if you’re ready, click on my wand for this week’s challenge!

Good Luck!

25 January 2022

Here are my mathematical conundrums for you this week:

1) Five angles in a hexagon are 110, 120, 130, 140 and 150 degrees. What is the sixth angle?

2) The reciprocal of a number is 1 divided by that number. For example, the reciprocal of 4 is 1/4, and the reciprocal of 2/5 is 5/2. Find all numbers that are quadruple their reciprocal.

Now when you have answered them, or done the best that you can, head over to the Answer tab to see if you are correct.

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

1) Five angles in a hexagon are 110, 120, 130, 140 and 150 degrees. What is the sixth angle?

Solution: The angles in a hexagon sum to 720, so the sixth angle is 720-(110+120+130+140+150)=720-650=70 degrees.

2) The reciprocal of a number is 1 divided by that number. For example, the reciprocal of 4 is 1/4, and the reciprocal of 2/5 is 5/2. Find all numbers that are quadruple their reciprocal

Solution: Let x be such a number. Then x = 4/x, so x^2 = 4, so x=2 or -2. So there are two such numbers: 2 and -2.

Well done if you worked out the right answers!

What is the smaller angle between the hour and minute hands of a clock at 09:40?

Solution: There are 30 degrees between adjacent numbers on the face of a clock, and the hour hand moves at 0.5 degrees per minute. So the required angle is 30+20=50 degrees.

One apple and two pears cost £1.36. Two apples and one pear cost £1.31. How much do nine apples and nine pears cost?

Solution: Three apples and three pairs cost £1.36+£1.31=£2.67, so nine apples and nine pears cost 3x£2.67=£8.01

In a triangle, the interior angles are in the ratio 3:4:5. What is the size of the largest angle?

Solution: There are 12 parts, and 180 degrees in total, so each part is worth 15 degrees. The largest angle corresponds to 5 parts, so is equal to 75 degrees.

What is the size of each interior angle in a regular 100-sided polygon?

Solution: The exterior angles of any polygon add to 360 degrees, so each exterior angle in a regular 100-sided polygon is 3.6 degrees, and hence each interior angle is equal to 180-3.6 = 176.4 degrees.

Two numbers have a sum of 2 and a product of -960. What is their difference?

Solution: The two numbers are 32 and -30, so their difference is 62

Two numbers have a sum of 1 and a product of -5/16. What is their difference?

Solution: The two numbers are 5/4 and -1/4, so their difference is 3/2 (or 1.5)

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

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

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

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

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?

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

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

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.

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?

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

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?

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

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

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

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

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

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

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

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

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

Find positive integers a and b, with a<b , so that 1/a + 1/b = 1/3

Solution: There are only a few numbers to try, as a can only equal 4, 5 or 6. The solution is a=4 and b=12

Positive integers a, b and c satisfy 1/a + 1/b + 1/c = 1/2 . What is the greatest possible value of c?

Solution: The greatest possible value of c is 42. This comes from 1/3 + 1/7 + 1/42 = (14+6+1)/42=21/42=1/2

The lowest common multiple (LCM) of 100 and N is equal to the highest common factor (HCF) of 100 and N. What is the value of N?

Solution: If the LCM and HCF of two numbers are equal, then the two numbers must be equal, so N=100

I’m thinking of a three-digit multiple of 4. If I reverse the digits, the new number is a multiple of 5. What is the largest possible value of my original number?

Solution: The number formed by the last two digits must be divisible by 4, and the first digit must be 5. So the greatest possible value of the original number is 596.

How many arrangements of the letters in the word ROTATOR read the same forwards and backwards?

Solution: The A must fall in the middle, and the first three letters must be O,R,T in some order. There are 3x2x1=6 ways to arrange the first three letters, and for each of these there is only one way to arrange the last three letters (they must mirror the beginning). Hence there are six valid arrangements in total.

If x is 2/3 of y, and z is 4/5 of x, what fraction of z is y?

Solution: z=(4/5)x=(4/5)(2/3)y=(8/15)y. So y=(15/8)z. That is, y is 15/8 of z.

What is the largest three-digit multiple of 6 that is still a multiple of 6 when its digits are reversed?

Solution A number is divisible by 6 precisely when it is even and its digits add to a multiple of 3. So we require the first and last digits to be even. The largest number satisfying the given conditions is 894.

How many minutes are there in one tenth of one third of one week?

Solution: There are 7x24x60 minutes in a week, so in one tenth of one third of this, there are 7x24x2=336

On a farm, the ratio of cows to pigs is 4:5 and the ratio of cows to sheep is 3:4. There are no other animals. What fraction of all the animals are cows?

Solution: The ratio of cows to pigs is 12:15, and the ratio of cows to sheep is 12:16. So the ratio of cows to pigs to sheep is 12:15:16, and the fraction of animals that are cows is 12/43

If really enjoy solving mathematical problems or if you’d just like a bit more practice, 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

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.

BBC Bitesize

Lots of fun problem solving questions for KS2.

https://www.bbc.co.uk/bitesize/subjects/z826n39

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.  