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I've come across this problem: $$\left\lfloor r + \frac{19}{100}\right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100}\right\rfloor + \dots + \left\lfloor r + \frac{91}{100}\right\rfloor = 546$$ $$\text{Find} \ \lfloor 100r\rfloor. \textit{(Source: AIME)}$$ Here is my work:

$\lfloor r + \frac{19}{100}\rfloor = r + \frac{19}{100} - \{r + \frac{19}{100}\}$, so the figure can be restated as $\{r + \frac{19}{100}\} = r + \frac{19}{100} + a - 546$, where $a = \lfloor r + \frac{20}{100}\rfloor + \lfloor r + \frac{21}{100}\rfloor + \dots + \lfloor r + \frac{91}{100}\rfloor$.

Because $\{r + \frac{19}{100}\}$ is the fractional part, $0 \le r + \frac{19}{100} + a - 546 < 1$, so after some more maniuplation, $545 + \frac{81}{100} \le r + a < 546 + \frac{81}{100}$. $a$ is an integer, so the fractional part of $r$ must be $\frac{81}{100}$.

$r = \lfloor r\rfloor + \{r\}$, so $\lfloor r + \frac{19}{100}\rfloor = \lfloor \lfloor r\rfloor + \frac{81}{100} + \frac{19}{100}\rfloor$ = $\lfloor \lfloor r\rfloor + 1\rfloor$. Because both of the terms inside that floor function are integers, it must equal $\lfloor r\rfloor + 1$. This same reasoning can be applied to each of the individual floor functions of the given figure's LHS, and they each turn out to be $\lfloor r\rfloor + 1$.

Therefore, $73 \lfloor r\rfloor + 73 = 546$, so $73\lfloor r\rfloor = 473$. However, this cannot be true unless there is no answer (which I assume is not the case), or unless I did something wrong in my process, because then $\lfloor r\rfloor$ is not an integer.

If you do see the solution, it would be really nice if you did not give the answer in your response! Instead, maybe some helpful hints or partial solutions would be preferred.

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    $\begingroup$ The line "...so the fractional part of $r$ must be $\frac {81}{100}$" is where I think I see an issue. Couldn't that inequality equally have any of the fractions in the interval $[\frac 9{100},\frac{81}{100}]$ based on applying that logic in the same manner to any of the other terms from the original? If you apply the correct term to your next line of reasoning, or else find the next-smallest multiple of $73$ below $473$ I think you'll have your answer... $\endgroup$
    – abiessu
    Mar 25 '20 at 17:40
  • $\begingroup$ Sorry, "next-largest" instead of "next-smallest"... Really, just begin with the "$73\lfloor r\rfloor+k=546$" logic, and deduce what $k$ has to be. $\endgroup$
    – abiessu
    Mar 25 '20 at 17:52
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    $\begingroup$ The average of the terms is about $7.5$ so each term must be $7$ or $8$. How many of each? $\endgroup$
    – saulspatz
    Mar 25 '20 at 17:58
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    $\begingroup$ Don't worry about the fractional part. And don't worry about $[r]$. Find the precise value where $[r + \frac k{100}]\ne [r+\frac {k+1}{100}]$. That is where $r + \frac {k}{100} < n \le r + \frac {k+1}{100}$ for some integer $n$. $\endgroup$
    – fleablood
    Mar 25 '20 at 18:23
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Don't worry about the fractional part. And don't worry about $[r]$. Find the precise value of $k$ where $[r + \frac k{100}]\ne [r+\frac {k+1}{100}]$. That is where $r + \frac {k}{100} < m \le r + \frac {k+1}{100}$ for some integer $m$.

====== my answer below ====

Well, what jumps at me is $0 < \frac k{100} < 1$ so all the $[r +\frac {k}{100}]$ are either one integer, call it $n$ or the next, $n+1$.

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So if $b$ of them equal $n+1$ and $(73 -b)$ of them equal $n$ we have $(73-b)n + b(n+1) = 73n + b =546$ where $0\le b < 73$.

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So as $546\equiv 35 \pmod {73}$ and $546= 7*73 + 35$ so $b=35$ and $n=7$.

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So we have (I'll have to be careful not to do a fencepost error... $91-35=56$ so....) $[r+\frac{19}{100}],....,[r+\frac{56}{100}] = 7$ and $[r+\frac {57}{100}],...,[r+\frac {91}{100}] = 8$.

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So $r + \frac {56}{100} < 8\le r+\frac {57}{100}$ so

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$100r + 56 < 800 \le 100r + 57$

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$100r < 744 \le 100r + 1$

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And $100r -1 < 743 \le 100r$ so $743 \le 100r < 744$

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So $[100r] = 743$.

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  • $\begingroup$ I solved the problem after user abieussu helped me realize that $\{r\}$ isn't necessarily equal to $\frac{81}{100}$, and your answer is wrong but very close. In your second to last step, you state this: "And 100r−<741≤100r so 741≤100r<742", which is the mistake, I believe. $\endgroup$
    – mpnm
    Mar 25 '20 at 19:00
  • $\begingroup$ If the third to last step is correct then my second to last ought to be. But arithmetic error can abound anywhere. $\endgroup$
    – fleablood
    Mar 25 '20 at 19:21
  • $\begingroup$ Oh, for gosh sake. I worried so much about making a fencepost error and figure $57$ was the correct fencepost marker.... and then immediately replaced it with $58$. So everything from then on is off by one$. $\endgroup$
    – fleablood
    Mar 25 '20 at 19:25
  • $\begingroup$ So .... corrected it but... might have made and error. $\endgroup$
    – fleablood
    Mar 25 '20 at 19:31
  • $\begingroup$ Argh... a fencepost error off by two... that's pretty annoying. $\endgroup$
    – fleablood
    Mar 25 '20 at 19:34
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$$ \left\lfloor r + \frac{19}{100}\right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100}\right\rfloor + \dots + \left\lfloor r + \frac{91}{100}\right\rfloor = 546$$

First, let $r = s - \dfrac{19}{100}$. Then the sum becomes

$$ \left\lfloor s + \frac{0}{100}\right\rfloor + \left\lfloor s + \frac{1}{100} \right\rfloor + \left\lfloor s + \frac{2}{100}\right\rfloor + \dots + \left\lfloor s + \frac{72}{100}\right\rfloor = 546$$

If $s$ were an integer, then you would get $73s = 546$, which has solution $s = 7 \dfrac{35}{73}$.

Noting that $7 \cdot 73 = 511$ and $8 \cdot 73 = 584$, we see that $7 < s < 8$. Now $546-511 = 35$. So where are you going to pick up that extra $35$?

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