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

```
7 years ago

68 Points
```										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.
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```
6 years ago
Rohan Chugh
20 Points
```										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.
```
3 years ago
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