# Inserting +/- into 123456789…

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:

1. A negative odd number will cancel a positive odd number, leaving an even number of odd numbers again.
2. 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?

• A negative odd number need not cancel out anything, after all, it may get summed together with an even number. see for example If I have $123456789$, I can set $123+45678+ 9$. In this way, the $1$ from $123$ is summed to $6$ and idoes not leave an even number... – 5xum Feb 11 '14 at 10:16
• In reply to Traklon: If 78 is base-10, yes. Not if it's base-9, where 78 = 71[b=10]. It's a generalization of the problem, so the formula is being expressed in base-b for base-b, i.e., not base-10 for base-9. – user127879 Feb 11 '14 at 10:34
• Yes, my mistake. I misread the question. – Traklon Feb 11 '14 at 10:41