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# REPOST 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 sing

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Let . When  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?)

May 24, 2018

### 2+0 Answers

#1
+102765
+1

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?)

Let t =  0.1212121212........

then 100t= 12.12121212.....

100t - t = 12

99t=12

t= 12/99

0.1212121212... = 12/99

Now do the same thing with 0.345345345.....

only since there are 3 digits repeating you will have to multiply by 1000.

Now if you understand what i have written you can finish it yourself.

May 24, 2018
#2
+102383
+1

0.12 (repetaing)  = 12/99 =   4/33

0.345 (repeating)  = 345/999  = 115/333

So

4/33  + 115/333  =

[4* 333 + 115 * 33] / [ 33 *333] =    1709 / 3663  =   .466557   (repeating)

So...the sum is  33

May 25, 2018