Find the number of solutions of the equation:[x]+[2x]+[4x]+[8x]+[16x]+[32x]=12345where [.] denotes the greatest integer function.

Dear   rishabh ,

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.

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Askiitians Expert

Ajit Singh Verma IITD

Let f(x) = [x] + [2x] + [4x] + [8x] + [16x] + [32x]. Obviously f is an increasing function (if x < y, then f(x) <= f(y).). f(196) = 12348. But if x is just under 196, then [x], [2x], [4x], [8x], [16x] and [32x] are all smaller by at least 1, so f(x) < 12342. Hence f never takes the value 12345.