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

68 Points
11 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.

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Ajit Singh Verma IITD

Rohan Chugh
20 Points
8 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.