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First or two articles about Fibonacci, written for students.

Three Neighbours

Age 7 to 11 Challenge Level:

There were many contributions to this activity. We were sent many from Monkfield Park Primary School, namely Kallum, Athur, Kiera and Oliver. Other good answers came from Maddie from Hurstpierpoint Prep School and Hannah, Sam and Gloria from Abington.

Dina from the Beit Shvidler Primary School
 in London sent in this good observation and a proof;

In the series of consecutive numbers, for any $3$ consecutive numbers, if the first number is $a$, the next is $a+1$, and the next is $a+2$. The sum is $3a+3$.

For the next group of $3$ numbers in the series, the $3$ consecutive numbers will be: $a+1, a+2, a+3$. This sum is $3a+6$.

This proves that a sum of $3$ consecutive numbers will always be $3$ more than the sum of the previous $3$ consecutive numbers in the series.


Amrit from Newton Farm Nursery, Infant and Junior School

Let the three consecutive numbers be $a - 1, a$, and $a + 1$.
Thus their sum is $3a$.
Thus the sum of any three consecutive numbers is a multiple of $3$.


 from St Joseph's Catholic School, Dorking
 showed a proof in another good way;

You take your three consecutive numbers and add them together. Then you look at your answer. All the answers are multiples of $3$ eg $3+4+5=12$

Then you will notice that the middle number, if timesed by $3$ will equal a multiple of $3$ eg $3+4+5=12, 4$x$3=12$

Another way to work it out, is to look at the outer numbers. Then take $1$ away from the last number and add it to the first number. You will notice that all the numbers are now the same. Add them together and the answer will be the same as in the other ways eg $3+4+5$ gives
$5-1=4$; $3+1=4$; $4+4+4=12$


Thank you for all your contributions.  Keep them coming in!