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Let $x$ be a real number. Denote by $[x]$ the integer part of $x$ and by $\{x\}$ the decimal part of $x$. Find the sum of all positive numbers satisfying $$25\{x\}+[x]=125.$$

I solved the question as:

To be noted first that since, $25$ and 125 are multiples of $5$, $[x]$ must be a multiple of $5$ too. Also, $[x]$ must be greater than $100$, since $[x]$ up to $100$ will also give an integer value for $\{x\}$. Further, $[x]$ must be up to $125$ only; if not, then $\{x\}$ will be negative. Considering all these factors, I substituted $[x]$ in the equation given as : $[x] = 105, 110, 115, 120$ and $125$, which gave me $\{x\}$ as $0.8, 0.6, 0.4, 0.2$ and $0.0$ respectively.

So, the real numbers come to : $105.8, 110.6, 115.4, 120.2$ and $125.0$.

So, the sum of all the numbers comes to : $577$.

Is this solution correct? Please advise. :-S

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  • $\begingroup$ The first sentence is already wrong, since {x} isn't an integer. $\endgroup$ – Kenny Lau May 28 '18 at 9:49
  • $\begingroup$ No, answer isn't correct. [x] isn't divisible by 5 $\endgroup$ – Love Invariants May 28 '18 at 9:49
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Hint: [x] and 125 are integers so 25{x} is also an integer.
Since {x} is less than 1
Equate 25{x} with all integers from 0 to 24.
Now for ,
25{x}=0 $\Rightarrow$ x+{x}=125
25{x}=1 $\Rightarrow$ x+{x}=124.04
25{x}=2 $\Rightarrow$ x+{x}=123.08
25{x}=3 $\Rightarrow$ x+{x}=122.12
25{x}=4 $\Rightarrow$ x+{x}=121.16
.
.
.
25{x}=24 $\Rightarrow$ x+{x}=101.96
Now adding all
$\Rightarrow$ 125+124.04+123.08 . . . 101.96
$\Rightarrow$ (125+124+123 . . . 101)+(0+0.04+0.08 . . . 0.96)
$\Rightarrow$ 2825+12
$\Rightarrow$ 2837

PS: You may recheck for the calculation part but the process is correct.

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  • $\begingroup$ Thank You @Love Invariants. That helped. Now, I got the sum as 2712. Is this answer correct now? :-S $\endgroup$ – Math Tise May 28 '18 at 10:19
  • $\begingroup$ @MathTise-Now the answer has been modified. $\endgroup$ – Love Invariants May 29 '18 at 14:03
  • $\begingroup$ You lost some zeros in the third to last line. $0.4$ should be $0.04$. I get $(125+101)25/2=2825$ for the first sum. There are $25$ terms that are all at least $101$, so $1695$ cannot be correct. $\endgroup$ – Ross Millikan May 29 '18 at 14:05
  • $\begingroup$ What about it now?? Is it OK? I have been going through brainfart. $\endgroup$ – Love Invariants May 29 '18 at 14:11
  • $\begingroup$ Thank You @LoveInvariants...greatly appreciated. $\endgroup$ – Math Tise Jun 12 '18 at 12:26

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