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2016 AMC 12B Problem 22

Problem 22 of 25HarderNumber TheoryArithmetic

For a certain positive integer nn less than 1000,1000, the decimal equivalent of 1n\dfrac1n is 0.abcdef‾,0.\overline{abcdef}, a repeating decimal of period 6,6, and the decimal equivalent of 1n+6\dfrac{1}{n+6} is 0.wxyz‾,0.\overline{wxyz}, a repeating decimal of period 4.4. In which interval does nn lie?

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Solution

Period 66 requires 106−1=33⋅7⋅11⋅13⋅3710^6-1=3^3\cdot7\cdot11\cdot13\cdot37 to be divisible by n.n. Period 44 requires 104−1=32⋅11⋅10110^4-1=3^2\cdot11\cdot101 to be divisible by n+6,n+6, while 102−1=32⋅1110^2-1=3^2\cdot11 is not divisible by n+6n+6 (else the period would be 11 or 22). Hence n+6n+6 is a multiple of 101.101. Since n+6n+6 also divides 32⋅11⋅1013^2\cdot11\cdot101 and is less than 1006,1006, the only possibilities are n+6=101,303,909,n+6=101,303,909, giving n=95,297,903.n=95,297,903. Only 297=33⋅11297=3^3\cdot11 divides 106−1,10^6-1, so n=297.n=297. Finally, 106≡1(mod297),10^6\equiv1\pmod{297}, while 102≢110^2\not\equiv1 and 103≢1(mod297),10^3\not\equiv1\pmod{297}, so its period is exactly 6.6. Also 303303 divides 104−110^4-1 but not 102−1,10^2-1, so the period of 1303\frac{1}{303} is exactly 4.4. Thus n=297n=297 lies in [201,400].[201,400]. Thus, the correct answer is B.
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Tagged: repeating decimal · multiplicative order · divisibility

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