Question

Let y=(0.\overline{12})+(0.\overline{345}). When y is written out as a repeating decimal, what is the sum of the digits in a single repeating period? (In other words, what is the sum of the digits covered by the repeat bar?)
Any help is greatly appreciated!

Answer 1

Answer:

33

Explanation:

The repeating digits of the sum will have a cycle of length 6 (the LCM of 2 and 3). There will be 3 cycles of the 2-digit repeat, and 2 cycles of the 3-digit repeat.

The sum of digits of the cycle of interest is the sum of the digits in 3 cycles of "12" and 2 cycles of "345", thus 3·(1+2) +2(3+4+5) = 9+24 = 33.

$0.\overline{12}+0.\overline{345}=0.\overline{121212}+0.\overline{345345}\\\\=0.\overline{466557}\\\\4+6+6+5+5+7=33$

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Another approach

You can look at the sum on a per-digit basis. The 2-digit repeat {12} has a sum of 1+2=3, or 3/2 on a per-digit basis. The 3-digit repeat {345} has a sum of 12, or 12/3=4 on a per-digit basis. The sum of the two repeating decimals will have a sum per digit of 3/2 + 4 = 5 1/2. The 6 repeating digits will then have a sum of 6×(5 1/2) = 33.

Answer 2

Answer:

33

Step-by-step explanation: