# Maths Wizard Challenge

## This Week’s Maths Wizard Problem…

Welcome back young maths wizards!

I have two more tricky mathematical conundrums for you to try this week!

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!

11 October 2021

Here are my mathematical conundrums for you this week:

1)  In a hexagon, what is the greatest possible number of interior right-angles?

2)  Martha thinks of a whole number. She squares it, then adds her original number, ending up with a prime. What number or numbers could she have been thinking of?

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)  In a hexagon, what is the greatest possible number of interior right-angles?

SolutionSix is impossible – the interior angle sum would then be 540 degrees instead of the required 720 degrees. The answer is five (make the hexagon look like a letter L).

2) Martha thinks of a whole number. She squares it, then adds her original number, ending up with a prime. What number or numbers could she have been thinking of?

SolutionWhether Martha begins with an odd number or an even number, the result will be even. But the only even prime is 2. It quickly becomes clear that large numbers (positive and negative) produce results that are also large; an analysis of numbers close to zero yields 1 and -2.

Well done if you worked out the right answers!

Bananas come in bunches of 3 (small bunch) or 5 (large bunch). Tim wants to purchase exactly 32 bananas; how many different combinations of bunches could he purchase?

Solution: The possible bunch combinations are 4 small and 4 large, or 9 small and 1 large, so there are just two combinations with a total of 32 bananas.

How many three-digit numbers contain the digit 3?

Solution: There are 9x10x10=900 three-digit numbers, of which 8x9x9=648 do not contain the digit 3, so the required total is 900-648=252.

On 27/09/21 the sum of the digits of the day (2+7) was equal to the number of the month (9). How many more times will this happen in 2021?

Solution: This will happen again on 19/10, 28/10 and 29/11; that is three more times.

Does there exist a number that is larger than its cube?

Solution:  There exist lots of numbers larger than their cube: any number between 0 and 1, and any number less than -1; so the answer is yes.

The reciprocal of a number is the result of dividing 1 by the number. For example, the reciprocal of 4 is 1/4. What is the reciprocal of 1/2 + 1/3?

Solution:   1/2 + 1/3 = 3/6 + 2/6 = 5/6, and the reciprocal of 5/6 is 6/5

A positive integer (whole number) n is divisible by 14 and 15. How many other positive integers are guaranteed to be factors of n?

Solution: Since n is divisible by 14 and 15, it must be divisible by the lowest common multiple of 14 and 15, which is 210. It must also be divisible by all the factors of 210: 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105, 210. So there are 16 factors of 210, of which we have already counted the factors 14 and 15; hence there are 14 other positive integers guaranteed to be factors of n.

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

The LCM of 10 and X is equal to ten times the HCF of 10 and X. Find all possible values of X.

Answer: X=1, 4, 25, or 100

Solution: If the LCM is ten times the HCF, one of the numbers contains one more power of 2 in its prime factorisation, and one of the numbers contains one more power of 5 in its prime factorisation. So the possible values of X are 1, 4, 25 or 100

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.  