Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
Make a set of numbers that use all the digits from 1 to 9, once and
once only. Add them up. The result is divisible by 9. Add each of
the digits in the new number. What is their sum? Now try some other
possibilities for yourself!
Can you explain the strategy for winning this game with any target?
There were two main approaches to this problem. One used number
properties combined with trial and improvement methods and other
focussed on the number properties.
Correct solutions were received from Sebastian Crane; Andrew
Yull, Chris Homes and Chris Wardale of Hethersett High School; Mary
Fortune of Birchwood Community High School; Roxane Foulser Piggott
of Oundle School; Matthew Causier of Queen Mary's Grammar School;
Jacob Graham, Jeongmin Lee and Alex Ward of the Mount School York
and Andrei Lazanu of school 205 Bucharest. Well done to each of
you. The solution below combines a mixture of the solutions
When the number x 1 x x x is multiplied by 417 this gives the
answer 9 x x x 0 5 7.
Find the missing digits, each of which is represented by an
The largest value for 9xxx057 is 9999057
The smallest value of 9xxx057 is 9000057
Dividing each of these by 417 gives answers that begin with a 2.
This means that the first digit of x1xxx must be 2.
We also know that the last digit must be 1, because it is the
only digit that multiplied by 7 gives 7 in the units column.
Next we can look at the 10's digit. This contributes to the 5 in
the tens column of the answer by:
being multiplied by the 7 in 417 and
being added to the product of the 1 in 417 and the last digit
(1) of the first number.
So we have:
(7x) +1$\times$1 = 5
This can only be the case if the last but one digit is a 2.
We can continue using long mulitplication - looking at the parts
of the multiplication that make the 0 in the hundreds column of the
We now have:
The zero in the hundreds column is formed from: 4$\times$1,
1$\times$2and7$\times$x + the 1 carried
from the tens column. These results combined must give a solution
with 0 in the units column:
4 + 2 + 7x + 1 = ?0
This will only work if x = 9
Now, we can write the entire multiplication: