Hardest maths questions - refactorisation and prime numbers 

As soon as I saw 65,I was instantly thinking of using that 11 with a 6, to get 66. That was a good exercise to help you move those numbers around.

Thank you for some fun recreational mathematics to keep my brain from getting too atrophied during lockdown. Here's another approach to this problem, though it doesn't show anything about factoring. The geometric mean, G, of a set of N numbers is the Nth root of their product, P. So you would expect the series of N numbers being sought to be clustered around the Nth root of P. P = 17,297,280 Try N = 5: 5th root = 28.02 suggesting 26,27,28,29,30 - wrong, also 29 cannot be constructed from given factors Try N = 4; 4th root = 64.49 suggesting 63,64,65,66 - Correct! Moreover N=3 or 2 do not work, showing that the 4 number sequence is unique. If you do not allow calculators this method can still help with approximate roots to indicate roughly where the correct sequence is.

A beautiful question

I have two A-levels in Maths, and paused the video mutliple times to try to get the answer before you, yet I still couldn't solve it. I tried to figure out how many factors of 2,3,5,7,11 and 13 were in the number and then look at the ranges between primes to see if I could find a range with the same. For some reason I incorrectly decided early on that there are 8 factors of 2 in the number, when there are actually only 7, so when I checked the range between 62 and 66, and then excluded 62 itself on the basis it was 2*31, and we didn't have a factor of 31 to use, I then concluded that the range did not have enough factors of two to be the correct one. Even when the video progressed and I was following your method and still pausing from time to time to get ahead of you revealing the answer I still couldn't get to the correct answer before you did because of my original error in miscounting the number of factors of two available to be used, which made me 'sure' that it couldn't be that range right up until the moment I saw a 7 and 9 were left after you divided up the 14 and used the resultant 2 together with 4 and 8 to get to 64. So yeah, I can completely understand why 0% of junior division AMC students (apparently that's Years 7 and 8, so 11 to 13 year olds) answered correctly.

I looked for gaps in the primes. Since 13 is the highest prime number that can occur in the prime factorization we can quickly rule out sequences which contain higher primes. For double digit numbers it is easy to see that you need a sequence of at least four consecutive numbers in order for the product to be big enough. The big enough gaps in the primes are between 23 and 29 (doesn't work since 25 has two factors of five), 31 and 37 (doesn't work since 34 = 17*2), 47 and 53 (rejected since 51 = 17*3), 53 and 59 (57 = 19*3), and finally 61 and 67 in which the right answer is contained.

Great explanation.

That's wonderful. How do you multiply those numbers buy combining circles?Is there a special app for that?

I would just multiply 15 times 20, and then subtract 15 twice.