If x=0.abcabc........(a repeating infinite decimal), what is the value

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 4 variables (x, a, b and c) and 1 equation (x = 0.abcabc….), E is most likely to be the answer. However, condition 1) includes 3 hidden equations (matching the values of the decimal places) and allows us to determine the values of all four variables as follows:
x = 1 – 0.123123123… = 0.876876876… = 0.abcabcabcabc…
Thus a = 8, b = 7, c = 6 and a + b + c = 8 + 7 + 6 = 21.
Condition 1) is sufficient.

Condition 2)
If x = 0.811811811…, then a = 8, b = 1, c = 1 and a + b + c = 10.
If x = 0.812812812…, then a = 8, b = 1, c = 2 and a + b + c = 11.
Since we don’t have a unique solution, condition 2) is not sufficient.


Therefore, A is the answer.
Answer: A

\(x = 0.\overline {abc} \,\,\,\,\left( {{\rm{common}}\,\,{\rm{russian}}\,\,{\rm{notation}}} \right)\)

\(? = a + b + c\)

\(\left( 1 \right)\,\,\,1 - x = 0.\overline {123} = {{123} \over {999}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,x = {{876} \over {999}} = 0.\overline {876} \,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,? = 8 + 7 + 6\)

\(\left( 2 \right)\,\,0.8 < x < 0.9\,\,\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,x = 0.\overline {801} \,\,\,\,\, \Rightarrow \,\,\,? = 8 + 0 + 1\,\, \hfill \cr \\
\,{\rm{Take}}\,\,x = 0.\overline {802} \,\,\,\,\, \Rightarrow \,\,\,? = 8 + 0 + 2 \hfill \cr} \right.\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

We need a unique value.

Statement 1)

1 - x = 0.123123123

Gives us a unique value for x. Sufficient.

Statement 2)

0.8 < x < 0.9

x could be anything so we cannot tell the value.

Insufficient.

Answer choice A

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