# Maths Wizard Challenge

## This Week’s Maths Wizard Problem…

**Hello young Maths Wizards!**

**Did you manage to answer my tricky maths conundrums last week? **

**Check the answers to see if you answered them correctly. If you sent in your answers do have a look at the leaderboards and see where you are. **

**Don’t forget, if you’d like some extra problem solving questions, look at the Further Resources tab, which includes some fun maths websites for you to try.**

**So click on my wand to see this week’s wizarding mathematical problems!**

**Good Luck!**

**23 November 2020**

**We have two new questions for you to test your mathematical thinking this week**

1)What is the smallest positive number divisible by 4, 5 and 6?

2)How many pairs of positive numbers (a,b), where a is less than b, have a lowest common multiple of 20?

*Check back next week to see if your answers are correct!*

Please submit your answers below and if correct you’ll name will be added to the leader board next week.

**Here are the answers to last week’s mathematical conundrums:**

1) 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?

* Answer:* £8.01

* 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**

2) One lemon and one lime cost 71p. One lemon and three limes cost £1.47. How much do three lemons and one lime cost?

* Answer:* £1.37

* Solution:* Two limes cost £1.47-£0.71=£0.76, so one lime costs £0.76/2=£0.38. Then one lemon costs £0.71-£0.38=£0.33, and so three lemons and one lime cost 3x£0.33+£0.38=

**£1.37**

**Well done if you worked out the right answers!**

**Maths Wizard Leaderboards**

Congratulations to everyone who correctly worked out the answers to my mathematical conundrums. Take a look at the boards below to see where you are this week.

Five points are awarded for each correct answer, so there are 10 possible points available to be won each week.

Leaderboards up to 23 November 2020

**Year 4**

**Year 5**

**Year 6**

1) Over a period of four days, Ellie eats several strawberries. Each day after the first day, she eats twice as many strawberries as she did the day before. In total she eats 90 strawberries.

How many strawberries did she eat on the last day?

* Answer:* 48

** Solution:** Ellie ate 6, then 12, then 24, then 48 strawberries across the four days. So she ate

**48**on the last day.

2) Over a period of four days, Taran eats lots of peas. Each day after the first day, he eats two thirds as many peas as he did the previous day. In total he eats 325 peas.

How many peas did he eat on the first day?

** Answer:** 135

** Solution:** Taran ate 135, then 90, then 60, then 40 peas across the four days. So he ate

**135**on the first day.

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

* Answer:* 75 degrees

* 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**.

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

* Answer*: 176.4 degrees

* 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**.

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

** Answer:** 50 degrees

** 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**

6) What is the smaller angle between the hour and minute hands of a clock at 16:59?

* Answer:* 155.5 degrees

* Solution:* There are 30 degrees between adjacent numbers on the face of a clock. The minute hands moves at 6 degrees per minute, and the hour hand moves at 0.5 degrees per minute. So, working backwards from 17:00, the required angle is 150+6-0.5=

**155.5 degrees**

7) How many positive numbers are factors of both 45 and 60?

**Answer**: 4

* Solution: *The common factors of 45 and 60 are precisely the factors of 15, namely 1, 3, 5 and 15. So the answer is

**4**.

8) How many positive numbers are factors of 100 but not 50?

* Answer*: 3

* Solutions: W*e require those factors of 100 that are divisible by 4, namely 4, 20 and 100, So the answer is

**3**

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

* Answer*: 62

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

**62**

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

* Answer:* 3/2 (or 5)

* Solution:* The two numbers are 5/4 and -1/4, so their difference is

**3/2**(or

**5**)

11) What is the smallest multiple of 10 whose digits sum to 10?

* Answer*: 190

* Solutions:* A number is divisible by 10 if and only if it ends with a 0. The answer is

**190**

12) What is the smallest multiple of 60 whose digits sum to 60?

** Answer**:

**79999980**

* Solution: *A number is divisible by 60 if & only if it is divisible by 3, 4 & 5. This requires the digits to have a sum divisible by 3, the number formed by the last two digits to be divisible by 4, and the last digit to be 0 or 5. The answer is

**79999980**

13) What is the smallest three-digit number that leaves a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5?

** Answer:** 119

** Solution:** The second condition requires a units digit of 4 or 9, and the first condition requires the number formed by the last two digits to be one less than a multiple of 4. So the answer is

**119**.

14) In how many ways can we pick three numbers from {1,2,3,4,5}, so that they are the sides of a triangle? (The order or the numbers doesn’t matter)

* Answer*: 3

** Solution**: We can pick any three numbers, so long as the sum of the smaller two sides is greater than the largest side. So the possibilities are (2,3,4), (2,4,5) and (3,4,5), and the answer is

**3**.

15) In how many ways can the letters in the word NOON be arranged?

* Answer*: 6

* Solutions*: The possible arrangements are NNOO, NONO, NOON, ONNO, ONON, OONN, so the answer is

**6**.

16) In how many ways can the letters in the word SCHOOL be arranged?

**Answer**: 360

** Solution:** Suppose all six letters were different. Then there would be 6 ways to choose the first letter, then 5 ways to choose the second letter, then 4 ways to choose the third letter, and so on. This yields 6x5x4x3x2x1=720 arrangements. But in the word SCHOOL, two of the letters are identical, so 720 has counted every possible arrangement twice, and the answer is

**360**.

**Why not take a look at some more tricky problem solving puzzles at the following websites. Just click on the link and it will take you straight there.**

**Junior Maths Challenge**

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

**Nrich Website / Twitter** (on twitter they post questions)

https://nrich.maths.org/primary

https://twitter.com/nrichmaths

**Times Tables Rockstars** (great fun reinforcing times table skills, for a small subscription of £6 pa)

**You may also like to take a look at Elastic Numbers, written by one of Hampton’s own Maths Teachers, Mr Griller.**

Available from most book shops and also online, *Elastic Numbers* is full of fun and challenging mathematical treats for the serious problem solvers among you!