For a certain positive integer n less than 1000, the decimal equivalen

Set .abcdef = X. Multiplying both sides by 10^6 results in

abcdef.abcdef = 10^6(X)

Subtracting .abcdef from left = abcdef

Subtracting this also from the right as X =

10^6(X) - X = X(10^6-1) = 999999X

So abcdef = 999999(.abcdef) so

.abcdef = abcdef/999999. Since .abcdef = 1/n,

n = 999999/abcdef.

By the same logic

n+6 = 9999/wxyz

Factoring 999999 = 3^3*7*11*13*37

Factoring 9999 = 3^2*11*101

So n is a factor of 999999 and n+6 is a factor of 9999

So which factor of 999999 is close to a factor of 9999 to home in on the answer ? This is annoying to figure out, but one factor of n+6 from above is

33, which divided into the numerator gives n+6 as potentially 303. This implies n=297

Can a number be divided into the factoring of 999999 to yield n=297?

3^3*11 is 297 and results from dividing the factoring by 7*13*37, so

yes, 297 is a possible value for n

So the answer that best fits is B.

1/297 = n= .003367 repeating and 1/303 = .0033 repeating


I don't take credit for this approach, only the explanation

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