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Stage: 2 Challenge Level: Challenge Level:2 Challenge Level:2

We had a very encouraging number of correct solutions, and from a good variety of countries around the word. These included: 


Libby, Harri, Elli, Robert and Joe, Charlotte and Karin, Holly and Leah, Olivia, Karman, George and Sarah, Jack, Sam, Keir, Alexander. Sophie, Laura, Romeo, Fatima, Nasrudin and Summer, Dana, Georgina, Alex and Megan, Ryan, Lara and Cordelia and Esther, Gavin, Eleanor.
Class 3/4 from Oak Hill 
Many from Moulsham Junior School
Many from St Kentigerns School
Many from Hethersett High School. 

Thinking of non-English schools:


Many from Jebel Ali Sch in Dubai,
Theo from International School Seychelles,
Krystof from Prague,
Tommy from International School in Sweden,
Lucas and Ollie from British School Brussels,
Alexander from Qatar
Insiya from Al Ameen Private School Dubai,
Primary 2/3 from Cumbrae Primary School Scotland.  
 

Liam and Ollie aged $8$ wrote;


 
To solve the problem we started by looking at the triangles. They were three because $12$ divided by $4 =3$.
The circles are two because in the third row we had two threes and two circles $6+4=10$.
The diamond and the hexagon then had to be $1$ or $5$.
The diamonds had to be worth $5$ because if you put a $5$ for the hexagon in the fourth column it wouldn't work because  then the total would be bigger than $5$. So then the hexagons had to be worth one.  

Sophie wrote a very clear description of her thinking and working; 


Firstly I worked out the value of the red triangles.  Column two had four red triangles that added up to twelve.  Twelve divided by four is three.
Next I worked out that the yellow circles are two. I did this by looking at row three, this had two triangles and two circles that added up to ten. The two triangles added up to six leaving four remaining which divided by two is two.
Next I worked out the value of the hexagon was one. Column four had to add up to five and had a circle and a hexagon already in it. As the circle equals two and the triangle equals three and the number had to be below five, that left four and one, it could only be one because there are two spaces left and the other two spaces already made three leaving one as the only available option.
Finally the blue diamond was five because the top row contains one of each shape and adds up to eleven, the triangle plus the circle plus the hexagon equals six. Eleven take away six is five.