Helen made the conjecture that "every multiple of six has more
factors than the two numbers either side of it". Is this conjecture
I start with a red, a green and a blue marble. I can trade any of
my marbles for two others, one of each colour. Can I end up with
five more blue marbles than red after a number of such trades?
I start with a red, a blue, a green and a yellow marble. I can
trade any of my marbles for three others, one of each colour. Can I
end up with exactly two marbles of each colour?
This solution came from Paul Jefferys.
We know that 2, 3, 5, 7 are strange. They are the only 1 digit strange numbers. All 2-digit strange numbers must be made up of these digits and must be prime. An exhaustive search reveals that they are 23, 37, 53, 73 as all other 2-digit numbers with these digits are not prime.
As no 2-digit strange numbers end in 2, and only 37 begins with a 3, we see that the only possible 3-digit number with 23 in it is 237, but this is not strange as it is a multiple of 3.
If we take 37, then we have to consider 237, 537, 737, 373, but 237 and 537 are multiples of 3, so we have 737 and 373, but then 737=11*67, so we are left with 373, which is prime.
Taking 53, the only possibility is 537 which is not prime.
From 73 there can only be 373 and 737, but 737 is not prime, so the only 3-digit strange number is 373.
Therefore there are NO 4-digit strange numbers as there is no 3-digit strange number ending 37. If there are no n-digit strange numbers then there cannot be any n+1 digit strange numbers as the shortened forms produced by removing the first or last digit of the n+1 digit number, which need to be strange, cannot be strange. So we have found all the strange numbers 2, 3, 5, 7, 23, 37, 53, 73,