MO C–I–1 2018

An unknown number is divisible by just four numbers from the set {6, 15, 20, 21, 70}. Determine which ones.

Result

x = (Correct answer is: 6,15,21,70) Wrong answer

Solution:

6=2⋅315=3⋅520=22⋅521=3⋅770=2⋅5⋅7LCM(6,15,20,21,70)=22⋅3⋅5⋅7=420  a=LCM(6,15,20,21,70)=420  x1=210 x2=630 x3=1050 x4=1470 x5=1890 x6=2310 x7=2730 ...  d(210)=1,2,3,5,6,7,10,14,15,21,30,35,42,70,... d(630)=1,2,3,5,6,7,9,10,14,15,18,21,30,35,42,45,63,70,...  LCM(6,15,21,70)=2⋅3⋅5⋅7=210  k=LCM(6,15,21,70)=210  x=6,15,21,706 = 2 \cdot 3 \\ 15 = 3 \cdot 5 \\ 20 = 2^2 \cdot 5 \\ 21 = 3 \cdot 7 \\ 70 = 2 \cdot 5 \cdot 7 \\ LCM(6, 15, 20, 21, 70) = 2^2 \cdot 3 \cdot 5 \cdot 7 = 420\\ \ \\ \ \\ a = LCM(6,15,20,21,70) = 420 \ \\ \ \\ x_{ 1 } = 210 \ \\ x_{ 2 } = 630 \ \\ x_{ 3 } = 1050 \ \\ x_{ 4 } = 1470 \ \\ x_{ 5 } = 1890 \ \\ x_{ 6 } = 2310 \ \\ x_{ 7 } = 2730 \ \\ ... \ \\ \ \\ d(210) = 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, ... \ \\ d(630) = 1, 2, 3, 5, 6, 7, 9, 10, 14, 15, 18, 21, 30, 35, 42, 45, 63, 70, ... \ \\ \ \\ LCM(6, 15, 21, 70) = 2 \cdot 3 \cdot 5 \cdot 7 = 210\\ \ \\ \ \\ k = LCM(6,15,21,70) = 210 \ \\ \ \\ x = 6,15,21,70

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