If a problem is too difficult to solve immediately, it can be valuable to solve a simpler problem instead, which can give us insight into the original problem. It typically won’t give us an answer, but can help us identify a suitable method.

We can simplify by replacing “difficult” numbers with simple round numbers to make calculation easier. Or we can replace a complex diagram with a simplified diagram that follows the same pattern.

**Examples of using a diagram to tackle a problem**

First we will read all three examples and have a quick think about them and then we will look at how simplifying the problem can help us with each one:

**Counting Squares Example**

How many squares (of any size) are there in the diagram below?

**Climbing Stairs Example**

Suzanne is climbing a flight of 10 stairs. She can either go from one stair to the next one (1 step), or she can miss out a stair and go up two at once (2 step). How many different ways can she go up the 10 stairs?

**Factors Example**

Find the smallest number that can be divided by all of the numbers from 1 to 20 without leaving a remainder.

**Worked Solutions to Examples**

**Counting Squares Example**

How many squares (of any size) are there in the diagram below?

**Solution**

We can consider progressively larger grids and look for a pattern. So we start with a 2×2 grid:

- This has:
- 1 large 2×2 square and
- 4 small 1×1 squares.

We then consider a 3×3 grid:

- This has:
- 1 large 3×3 square,
- 4 medium 2×2 squares and
- 9 small 1×1 squares.

And finally a 4×4 grid:

- This has:
- 1 4×4 square,
- 4 3×3 squares,
- 9 2×2 squares.

You can probably notice a pattern from these examples. If you apply this pattern to the question, how many squares will there be in total?

**Climbing Stairs Example**

Suzanne is climbing a flight of 10 stairs. She can either go from one stair to the next one (1 step), or she can miss out a stair and go up two at once (2 step). How many different ways can she go up the 10 stairs?

**Solution**

Imagine first a staircase with only 1 stair, for which there will be only 1 way of climbing it, then one with 2 stairs for which their will be 2 ways, then one with 3 stairs for which there will be 3 ways. If we continue in this manner until we are considering a staircase with 10 squares, we find the following:

Number of Stairs | Number of Ways to Climb Them |

1 | 1 |

2 | 2 |

3 | 3 |

4 | 5 |

5 | 8 |

If we look carefully at the second column we may be able to recognise that it is a known sequence, the Fibonacci sequence. We could test this for 6 stairs to check whether 13 seems a suitable answer, and then we can extend the sequence up to 10 stairs.

**Factors Example**

Find the smallest number that can be divided by all of the numbers from 1 to 20 without leaving a remainder.

**Solution**

Suppose it only had to be dividable by the number from 1 to 3. Then clearly this would have to be 6, as there is no smaller number which has factors 1, 2 and 3.

What if it was 1 to 4? Well, multiplying these numbers together gives 1 x 2 x 3 x 4 = 24. But this is not the smallest number with all these as factors. Because the 4 already has one 2 as a factor, we don’t need to include that 2 twice and so we can take 2 x 3 x 2 = 12.

What if it was 1 to 5? Well we found above that 12 is divisible by 1 to 4. But it isn’t divisible by 5, so we will have to multiply this by 5 to give 60. You many notice that it matters whether the number is a prime or a composite number.

What if it was 1 to 6? Well, we already found that 60 is divisible by 1 to 5, and it is also divisible by 6, so this is the smallest number.

What if it was 1 to 7? As 7 is a prime we must multiply this to give 60 x 7 = 420.

For 1 to 8 we only need one extra 2 because our number is already divisible by 4, so we have 60 x 7 x 2 = 840

For 1 to 9 we only need one extra 3 because our number is already divisible by 3, so we have 60 x 7 x 2 x 8 = 2520.

For 1 to 10 we notice that 2520 is already divisible by 10. So written as a product of primes, our number is now 2 x 2 x 2 x 3 x 3 x 5 x 7

We then carry on with the same logic through the numbers from 11 to 20 to find out what our “smallest possible number” must be multiplied by in order to have these as factors, as shown below.

11 | Need to multiply by it, it is prime | 16 | We already have 8 (i.e. 2 x 2 x 2 above), but need an extra 2 |

12 | Already included (i.e. 2 x 2 x 3 above) | 17 | Need to multiply by it, it is prime |

13 | Need to multiply by it, it is prime | 18 | Already included (i.e. 2 x 2 x 3 x 3 above) |

14 | Already included (i.e. 2 x 7 above) | 19 | Need to multiply by it, it is prime |

15 | Already included (i.e. 3 x 5 above) | 20 | Already included (i.e. 2 x 2 x 5 above) |

So from the table we have additional factors, of 11, 13, 2, 17 and 19 to multiply by, biving us the final product of 2 x 2 x 2 x 2 x 3 x 3 x 5 x 7 x 11 x 13 x 17 x 19. This equals 232,792,560.

**17 Questions of Increasing Difficulty**

**Worked Solutions to Questions**