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

## 2 Answers

10 years ago

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.

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

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Askiitians Expert

Ajit Singh Verma IITD

7 years ago
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.

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