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, [π] = 3 and [-5/ 2] = -3. Show that the equation

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

has no real solution.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to Catherine Diễm)

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 now have

63i + [f] + [2f] + [4f] + [8f] + [16f] + [32f] = 63i + [2f] + [4f] + [8f] + [16f] + [32f] = 12345 = 63 × 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. Therefore, equation (*) is not possible, and there is no f that can satisfy the equation in the problem, and thus there is no x.

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