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 2x, 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 x 13 = 360,360. (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.
