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02 Jun 2022, 01:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
31% (01:52) correct
69%
(01:43)
wrong
based on 52
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For a certain positive integer n less than 1000, the decimal equivalent of \(\frac{1}{n}\) is \(0.\overline{abcdef}\), a repeating decimal of period of 6, and the decimal equivalent of \(\frac{1}{n+6}\) is \(0.\overline{wxyz}\), a repeating decimal of period 4. In which interval does n lie?
(A) [1,200]
(B) [201,400]
(C) [401,600]
(D) [601,800]
(E) [801,999]
Are You Up For the Challenge: 700 Level Questions
(A) [1,200]
(B) [201,400]
(C) [401,600]
(D) [601,800]
(E) [801,999]
Are You Up For the Challenge: 700 Level Questions
02 Jun 2022, 02:18
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Whoever can solve this question within minutes must cruise on to ace 51 on GMAT Quant.
For the rest of us, learn to quickly identify questions at this difficulty level and skip it right away to allow the test engine to properly recalibrate its overinflated expectations on our math skills.
For the rest of us, learn to quickly identify questions at this difficulty level and skip it right away to allow the test engine to properly recalibrate its overinflated expectations on our math skills.
12 Jun 2022, 12:26
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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
Posted from my mobile device
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
Posted from my mobile device
