### Pebbles

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!

### GOT IT Now

For this challenge, you'll need to play Got It! Can you explain the strategy for winning this game with any target?

# X Marks the Spot

##### Stage: 3 Challenge Level:

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 received.

### The Problem

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 "x".

### A Solution

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.

 21xx1 417 ------------ 9xxx057

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 answer.

We now have:

 21x21 417 ----------- 9xxx057

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:

 21921 x 417 ------------ 9141057