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Bài giải bài toán vô nghiệm

[ Friday, 05 February, 2010 ]

Problem 1 of the Canadian Mathematical Olympiad 1981

For any real number t, denote by [t] the greatest integer which is less than or equal to t. For example: [8] = 8, [pi] = 3 and [-5/ 2] = -3. Show that the equation

[x] + [2x] + [4x] + [8x] + [16x] + [32x] = 12345

has no real solution.

Solution by Vo Duc Dien

Let x = i + f  where i is the integer part or integral part and f the fractional part of x.

We have f < 1, and

[x] + [2x] + [4x] + [8x] + [16x] + [32x] = 63i + [f] + [2f] + [4f] + [8f] + [16f] + [32f]

since f < 1,  [f] = 0,         and we have

63i + [f] + [2f] + [4f] + [8f] + [16f] + [32f]  =  63i + [2f] + [4f] + [8f] + [16f] + [32f]

= 12345 = 63 x 195 + 60

So     i = 195                  and    [2f] + [4f] + [8f] + [16f] + [32f]  =  60          (*)

Since  max[nf] = n – 1, the maximum value of  [2f] + [4f] + [8f] + [16f] + [32f] =

1 + 3 + 7 + 15 + 31 = 57. Equation (*) is impossible. Therefore, there is no f that can satisfy the equation in the problem, and thus there is no x.

Extension of problem:     Once can change the number 12345 to 12344 or 12343 and the problem is still valid.

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