Community Primary School
It was lovely to see we had over 20 unique submissions for the puzzle this week, 13 of which were spot on. Congratulations to those who got the correct answer (see the solution in the Past Puzzles section below), and thank you to all who took the time to give the puzzle your best.
It's half-term, so there will be no new puzzle next Friday. Instead, we'll publish 2 puzzles today to keep you challenged through the week. Keep your eyes peeled for when these go live - we'll alert you through the app as usual.
Each week we will publish a new puzzle or problem for you to solve. The puzzles won't always have a maths theme: sometimes you'll be faced with word or picture problems.
You're welcome to get help from friends and family, or use the internet to help. Once you have worked out the answer, fill in the form below with your answer, name and class.
One winner will be randomly drawn from all entries received and they will be presented with a certificate in assembly at the end of the week.
You can only submit your answer once - multiple entries will be removed from the draw.
Deadline for entries is Thursday 28th February. Good luck!
Puzzler #5 - Puzzler of the Week - Ebony Hitchen (Y4)
Q. In this partly completed pyramid, each rectangle is to be filled with the sum of the two numbers in the rectangles immediately below it.
What number should go in the red rectangle?
(There's no trick here - only maths!)
The answer is 3.
From top to bottom: 105; 58, 47; 31, 27, 20; 17, 14, 13, 7; 12, 5, 9, 4, 3
Q. In the sum below, the digits have been replaced by shapes. Identical shapes have the same value. What is the value of the square?
The answer is 6.
The sum consists of an unknown value of ones + the same unknown value of ones + an unknown values of tens and ones, resulting in an unknown value of hundreds, tens and ones. With each digit in the result being identical, the highest possible result when adding the described values is 111. Using similar logic, the only possible value of the circles is 99. Once you have realised this, you can calculate the value of the square by inverse operation:
111 = 99 + 2 squares, so 2 squares = 12, and so 1 square = 6
A tricky puzzle, the hardest one yet. Well done if you got the correct answer.
Q. The matchsticks in the picture are arranged to form a number sentence in Roman numerals, but obviously six minus four is not nine. Which matchstick (labelled 1 to 12) would you need to move in order to make the number sentence correct?
Image taken from parallel.org.uk - older pupils may wish to check out this great (challenging!) maths activity site.
Move matchstick 3 so that it crosses matchstick 4 to make a plus sign. Then we have 5 + 4 = 9.
In the image below, what is the number of the parking space taken by the car?
An easy one if you spotted the picture was upside-down! The answer is 87, because if you look it at from the driver's point of view, then the numbers read 86 (not 98), 87, 88, 89, 90, 91 (not 16).
The diagram below shows a rectangle placed on a grid of 1 cm × 1 cm squares.
What is the area of the rectangle in cm2?
The total area of the 8 × 8 grid in cm2 is 82 = 64.
The two larger triangles in the top left and bottom right corners of the grid make up a 5 × 5 square with area 25 cm2.
The two smaller triangles in the other corners of the grid make up a 3 × 3 square with area 9 cm2.
So the area of the rectangle, in cm2, 64 - 25 - 9 = 30.