Helen's Conjecture

Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?

Marbles

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?

More Marbles

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?

Where Can We Visit?

Stage: 3 Challenge Level:

Fantastic responses to this problem, Benjamin and George discovered this:

We started with 42 and using the rule multiply by 2 and subtract by 5.
We discovered when any number with the factor of 2 is doubled (multiplied by 2) it gives a units digit of 4, 8, 6, and back to 2 which we started with. Although you can get round this by subtracting 5 it only then takes you to 7, 3 1 and 9 which with the 10 by 10 grid it becomes impossible to numbers with the factors or multiples of 5 e.g. 5, 10 15...

Lydia and Jess explain their findings:

We can not visit numbers which are multiples of 5 (those ending in a 5/0), 99 or 97. We could not visit multiples of 5, as starting from 42, multiplying by 2 or subtracting by 5 would not get us to a multiple of 5. Multiplying 42 by 2 results in 84, you cannot do this again as the result would be above the number 100; however many times you minus 5 you wold only visit numbers ending in 7 or 2. We could not visit 97 or 99 as they are not even meaning we can not double any number to get there. both of these are closer to 100 than 5 meaning we can not subtract 5 to there.

Lyman Shen explored other combinations of multiplying and subtracting:

Solution for $\times$ 2 and -5:
First I started with 1. By multiplying by two I can go to 2, 4, 8, 16, 32, 64. And I go down 64 from -5 all the way to 9 (4 would do the same thing again). And then I multiply by 2 again, and get 9, 18, 36, 72. I can carry this on to get every number, except I can't get any multiple of 5's. Also I can't make 97, and 99. I can't get them because they are odd numbers and can't get it by -5. I can get 98 from $\times$ 2 from 49, which could be get from -5 form numbers end with 9 and 4.

Solution for $\times$ 3 and -5:
By working similarly to before, I can find all the numbers except for the multiple of 5's because no numbers can make it to 5, 10, unless I start with a multiple of 5. I also can't get 92, 97, 98. I can't get them because they aren't multiple of 3's and can't get it from -5.

Solution for $\times$ 4 and -5:
This is time the numbers that can be reached depend on the numbers started with. When the starting number ends with 1, 4, 6, 9, there are no way to get any number that ends differently ($1 \times 4 = 4, 4 \times 4 = 16, 6 \times 4 = 24, 9 \times 4 = 36$). So we can reach everything ending in 1, 4, 6, 9 except the large numbers 94, 99. I can't get those number because it is not divisible from 4 and I can't get them from -5, which I can do to 96.
When the starting number ends with 2, 3, 7, 8, there are no way to get any number that ends differently ( $2\times 4 = 8, 3\times 4 = 12, 7\times 4 = 28, 8\times 4 = 32$). So I can reach anything ending in these digits, apart from 97. I can't get it because it isn't divisible from 4 and can't get it from -5.
When the start number ends with 5 or 0, there are no way to get any number that ends in anything other than 0 and 5.

Solution for $\times$ 5 and -5:
This reaches every multiple of 5, and the number you start with.
The reason is $5\times x = 5x$, which have to end with 0 or 5. No more exceptions after that.

Solution for $\times$ 6 and -5:
This is another one that is dependant on the starting number.
If it ends with 1 or 6, you can't get any numbers that don't end with 1 or 6 ($1 \times 6 = 6, 6 \times 6 = 36$).
If it ends with 0 or 5, you can't get any numbers that don't ends with 0 or 5 ($0 \times 6 = 0, 5 \times 6 = 30$). The ones you can't reach 95 and 100 ($15 \times 6 = 90$).
If it ends with 4 or 9: The same thing ($4 \times 6 = 24, 9 \times 6 = 54$). There are exceptions 99, 94, 89 ($14 \times 6 = 84$).
If the stating number ends with 3 or 8. $3 \times 6 = 18$, $8 \times 6 = 48$ so again you can reach all numbers ending in 3 or 8, but missing 98, 93, 88, 83 as the highest that can be reached is $13 \times 6 = 78$