Average digit sum – 3

Number Theory Level 4

Balanced ternary is a number system that uses three digit corresponding to the place values of − 1 , 0 , 1 -1, 0, 1 , in contrast to ternary that uses 0 , 1 , 2 0, 1, 2 .

The first numbers written in balanced ternary (using the "digits" − , 0 , + -, 0, + ) are

decimal balanced ternary
0 10 0_{10} 000 000
1 10 1_{10} 00 + 00+
2 10 2_{10} 0 + − 0+-
3 10 3_{10} 0 + 0 0+0
4 10 4_{10} 0 + + 0++
5 10 5_{10} + − − +--
6 10 6_{10} + − 0 +-0
7 10 7_{10} + − + +-+
8 10 8_{10} + 0 − +0-
9 10 9_{10} + 00 +00
⋮ \vdots ⋮ \vdots

The digit sum of a balanced ternary number can be calculated just like the normal digit sum, with the exception that − - corresponds to − 1 -1 , so subtracting 1 in the digit sum. For example, the digit sum of 5 = + − − 5 = +-- is + 1 − 1 − 1 = − 1 +1-1-1 = -1 . This shows that the digit sum can also be negative.

Now, what is the average digit sum of all integers between 1 and 121 (both inclusive) when they're written in balanced ternary?

Other parts


The answer is 1.

Henry U by Henry U , 17, Germany

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1 solution

David Vreken
Dec 18, 2018

The first balanced ternary number (for the number 1) is +, for a digit sum of 1.

The next 3 balanced ternary numbers (for the numbers 2-4) are +-, +0, and ++. There are an equal number of +, 0, and - symbols in the last digit place, so those have no effect on the net digit sum, but the first digit place are all +'s, for an additional 3 to the total digit sum.

The next 9 balanced ternary numbers (for the numbers 5-13) are +--, +-0, +-+, +0-, +00, +0+, ++-, ++0, and +++. There are an equal number of +, 0, and - symbols in the last and second-to-last digit place, so those have no effect on the net digit sum, but the first digit place are all +'s, for an additional 9 to the total digit sum.

The same logic can be repeated for each consecutive 3 n 3^n grouping of balanced ternary numbers, and we find that we add an additional 3 n 3^n to the total digit sum each time. Since 121 = 1 + 3 + 9 + 27 + 81 121 = 1 + 3 + 9 + 27 + 81 , the total digit sum would also be 1 + 3 + 9 + 27 + 81 = 121 1 + 3 + 9 + 27 + 81 = 121 , which makes the average digit sum of the first 121 numbers 121 121 = 1 \frac{121}{121} = \boxed{1}

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