Maths Problem (1)

Here is the first question which I reckon an A level student would
find fun to explore but I have no idea how hard the proof is.

If I take an n digit number I can create another n digit number by
taking as each digit M in the new number the difference between digit M
and digit M+1 in the old number and the putting the difference between
the digit 1 and n of the first number on the end.

With young kids you can just play with this with say 4 digit numbers
which all end up as zero: 3972 gives 6251 gives 4345 gives 1111
gives 0000.

It is self-correcting. If you slip up you still get to zero.

With older kids you can try with other numbers of digits.

Hence 896039 gives
136361 gives 233320 gives 100122 gives 101101 etc.

If I repeat this process I may reach zero: 00000 gives 00000 or for
some n I may reach another stable solution sequence like 110 gives 011
etc. Since the numbers are finite and the steps repeatable only cycles
or stopped end points are possible.

Experimenting seems to show that if n is an exact power of 2 then
everything resolves to zero but for n not an exact power of two other
stable sequence solutions exist. Can anyone prove this conjecture right
or wrong?

Maths Problem (2)

This is a serious maths problem which drives investments in retail networks. I call it the

Retail Network Attraction/Convenience problem.

This problem is laid down as a challenge to network theorists. It is a simple representation of an incredibly important real problem. It was fairly well formulated in the retail oil company where I used to work but despite asking around we never found anyone who had made any progress with it.

Suppose customers are identical and uniformly distribution over some space (R2 is fine but in reality some convenience metric applies). Say the customer density is D. Suppose retail outlets are randomly distributed throughout the space, average density d, and I own X% of them. If customers always buy from their nearest outlet then my market share is X%. But what if I can change things (price, site facilities, TV advertising) so that customers will prefer my outlets to my competitors and be prepared to travel an additional x m for my outlet, so that they will buy from my site if my site is nearest or no more than x further than any competing site: what is my new market share? Should I invest in new sites or improve the ones I have?

All the other elements of the commercial problem are understood. Considerable good quality research (commercially secret) has been done on site and brand attractiveness and the most price effective way to achieve it (i.e. x is known as a function of investment: this is typically done by conjoint multi-linear regression)). The economic effect of changing price at a given market share is clear. The cost saving from closing sites is well known. But without cracking the network problem the basic problem of where to put your investment (new sites, overall brand, site by site improvement) as a function of d, D and X is unknown. Lots of other issues (is the whole network over saturated? Can I close sites to save operational cost without much loss of market share provide my attractiveness is at least x more than competitors etc.) hangs on this analysis. What if I close my smallest sites etc.?

Whatever empirical data on effects of site investments or price changes exists to the best of my knowledge is tarnished by being at a single market share, by competitors not standing still and by customer memory effects (steady state make take a year; e.g. customers find it famously difficult to compare prices). So real investment strategy would be steered by insight into this problem.

At least in broad terms a solution should show that in very sparse networks with low market share you should always build sites. In very dense networks you should always invest in attractiveness. But all in all I think this is a nice research problem.

This is the simplest version. The main change going to the next level up is changing from assuming a single attraction/convenience trade off for a customer, to using a distribution.

Look forward to answers. Any real progress should in my view be properly published in peer reviewed journals but I am happy putting Qs or clarifications on this site or helping.

Ideas for teachers.

Cyclic number

An interesting number.

142857 is a cyclic number. When multiplied by 1,2,3,4,5,6 is just permutes its digits. Why?

Because 142857/999999 = 1/7. Or put another way 1/7 = 0.142857142857142857.....

Whenever a fraction 1/m is calculated by long division, if the remainder runs through every possibility before repeating you get a cyclic number. In practice this requires m to be a prime number.

You can get the less able kids to work out the 142857 x 1, 2, 3, 4, 5, 6, 7 and the abler ones can work on why and trying to find some others

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More problems will be added when I have time

If you have any comments please email: | maths | @ | catesfamily.org.uk |

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