I'm looking at a generalization of the problem of inserting + and/or - into the blocks $123456789$ and $987654321$ to create a formula for $100$, like this:
$$123 - 45 - 67 + 89 = 100$$
$$9 - 8 + 7 + 65 - 4 + 32 - 1 = 100$$
Generalizing the problem in base-b, one needs to insert +/- into the block $[12345\dots b-1]$ to create a formula for $b^2$ (expressed in base-b). Such formulae exist in base-$11$ and base-$12$:
$$123 + 45 + 6 + 7 - 89 + A = 100[b=11]$$
$$146 + 49 + 6 + 7 - 97 + 10 = 121$$
$$123 + 4 + 56 + 7 - 89 - A + B = 100[b=12]$$
$$171 + 4 + 66 + 7 - 105 - 10 + 11 = 144$$
But I have found no such formulae in base-$9$, base-$13$ and base-$17$. I assume this will be true of all base-[4n+1], because the block of integers in such bases will consist of an even number of odd numbers and an even number of even numbers. E.g., for base-$9$, the block is $12345678$, i.e. odds are $1,3,5,7$ and evens are $2,4,6,8$. Therefore one can't find a +/- formula in base-$[4n+1]$ that sums to an odd number, because:
- A negative odd number will cancel a positive odd number, leaving an even number of odd numbers again.
- Numbers in base-$[4n+1]$ with more than one digit, like $12$ and $456$, will be odd/even in such a way as to leave the addition of odd numbers unaffected.
My questions are:
Is this proof correct?
Is there a simpler way to express it?