Earlier today I set you the following three problems.

**Solution: 64**

If we call the ‘size’ of a triangle the number of individual areas it contains, there are

7 triangles of size 1

14 triangles of size 2

10 triangles of size 3

13 triangles of size 4

6 triangles of size 6

6 triangles of size 8

3 triangles of size 9

4 triangles of size 12

1 triangle of size 16

Add them up at that makes 64 triangles in total.

**Solution: 25 per cent (or 1/4)**

Every ant can move in two possible directions, which makes 2³ = 8 possible sets of outcomes, all of which are equally likely.

Only two of the outcomes have no collisions: when all ants are moving clockwise or anticlockwise. This means the probability is 2/8 = 1/4, or a 25 per cent chance.

**Solution: The statements ‘It is divisible by 16’ and ‘It is divisible by 17’**

Let N be the large integer that is the subject of the question.

We are looking for the two incorrect statements, which we are told are consecutive.

The first statement, that N is divisible by 1, must be true because every whole number is divisible by 1.

The second statement, that N is divisible by 2, must also be true because if it was false, then all the even numbered statements would also be false, since N would not be divisible by 4, 6, 8 and so on, which would mean there are 15 incorrect statements, contradicting the premise.

Using the same argument, the first 15 statements must also all be true. If N were not divisible by *x, *where *x *is* *between 1 to 15, then N would not be divisible by 2*x*, so you would have at least two incorrect but nonconsecutive statements.

We conclude that the two incorrect statements must concern the numbers between 16 and 30.

We know that N is divisible by all numbers from 2 to 15. We can express these numbers in terms of their prime factors:

2

3

2 x 2

5

2 x 3

7

2 x 2 x 2

3 x 3

2 x 5

11

2 x 2 x 3

13

2 x 7

3 x 5

N must also be divisible by the **lowest common multiple** of all these numbers, which is 2³ × 3² × 5 × 7 × 11 = 27720. (8 gives us three copies of 2, and 9 gives us two copies of 3.)

This number is divisible by 18 (= 2 × 3²), 20 (= 2² × 5), 21 (= 3 × 7), 22, 24, 26, 28 and 30, so N is also divisible by these numbers.

We are left with just seven possibilities for incorrect factors:

The prime numbers 17, 19, 23 and 29

The powers 16 = 2⁴, 25 = 5² and 27 = 3³. Each of these powers contains more copies of the respective prime factors than in the prime factorisation above.

Of these candidates, there is just one pair of consecutive numbers: 16 and 17.

I hope you enjoyed today’s puzzles. I’ll be back in two weeks.

*I set a puzzle here every two weeks on a Monday. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.*

*Thanks to Philipp Legner for today’s puzzles. Check out his amazing website Mathigon.org.*