# Divide $40$ to $4$ parts such that every number from $1-40$ can be realized

How can you divide $40$ to $4$ parts such that every number from $1-40$ can be realized just by adding or subtracting those $4$ parts?

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See the balanced ternary number system. –  Marc van Leeuwen Dec 13 '12 at 12:41

All integers from 1 to 40 can be expressed by adding or subtracting 1,3,9 and 27 in such a way that each number is used at most once, and it should be either added or subtracted.

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You mean, "each number is used at most once". –  TonyK Dec 13 '12 at 13:25
Oops. That's true. –  Aneesh Karthik C Dec 29 '12 at 7:04

My answer is the same as Aneesh, but I'll try to give some informal intuition/justification of the answer. Think of it this way: What numbers would you have needed if you could only add them atmost once? So, you could say that the coefficients of each of these numbers would be either 0 or 1. This translates to the binary number system, and you would require 1, 2, 4, 8, 16, 32 to generate all numbers from 1 to 40 (all powers of 2 less than 40).

Now look at your question. Going on similar lines, the appropriate coefficients would be -1, 0 and 1. You can treat it as a base-3 (ternary) number system, and you would need all the powers of 3 (less than 40) to generate all numbers from 1-40, namely: 1, 3, 9 and 27.

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