AIME 1991 · 第 6 题

Solution 1

There are $91 - 19 + 1 = 73$ numbers in the sequence. Since the terms of the sequence can be at most $1$ apart, all of the numbers in the sequence can take one of two possible values. Since $\frac{546}{73} = 7 R 35$, the values of each of the terms of the sequence must be either $7$ or $8$. As the remainder is $35$, $8$ must take on $35$ of the values, with $7$ being the value of the remaining $73 - 35 = 38$ numbers. The 39th number is $\lfloor r+\frac{19 + 39 - 1}{100}\rfloor= \lfloor r+\frac{57}{100}\rfloor$, which is also the first term of this sequence with a value of $8$, so $8 \le r + \frac{57}{100} < 8.01$. Solving shows that $\frac{743}{100} \le r < \frac{744}{100}$, so $743\le 100r < 744$, and $\lfloor 100r \rfloor = \boxed{743}$.

Solution 2 (Faster)

Recall by Hermite's Identity that $\lfloor x\rfloor +\lfloor x+\frac{1}{n}\rfloor +...+\lfloor x+\frac{n-1}{n}\rfloor = \lfloor nx\rfloor$ for positive integers $n$, and real $x$. Similar to above, we quickly observe that the last 35 take the value of 8, and remaining first ones take a value of 7. So, $\lfloor r\rfloor \le ...\le \lfloor r+\frac{18}{100}\rfloor \le 7$ and $\lfloor r+1\rfloor \ge ...\ge \lfloor r+\frac{92}{100}\rfloor \ge 8$. We can see that $\lfloor r\rfloor +1=\lfloor r+1\rfloor$. Because $\lfloor r\rfloor$ is at most 7, and $\lfloor r+1\rfloor$ is at least 8, we can clearly see their values are $7$ and $8$ respectively. So, $\lfloor r\rfloor = ... = \lfloor r+\frac{18}{100}\rfloor = 7$, and $\lfloor r+\frac{92}{100}\rfloor = ...= \lfloor r+1\rfloor = 8$. Since there are 19 terms in the former equation and 8 terms in the latter, our answer is $\lfloor nx\rfloor = 546+19\cdot 7+8\cdot 8=\boxed{743}$

~ minor edit by arjunkhokhar

Note

In the contest, you would just observe this mentally, and then calculate $546+19\cdot 7+8\cdot 8= 743$, hence the speed at which one can carry out this solution.

Solution 3 (Algebra)

Notice how all terms in this series is in the form $\lfloor r + \frac{k}{100} \rfloor$, where $19 \le k \le 91$. This then implies that there are a total of 73 terms in this series.

If every term is the same, i.e., $73 \lfloor r + \frac{19}{100} \rfloor = 546$, then $\lfloor r + \frac{19}{100} \rfloor = \frac{546}{7} \approx 7$. However, because this isn't exactly 7, this implies that the values in the series are mixed between 7 and 8, so we need to do a bit more work to find out the value of $k$ such that $\lfloor r + \frac{k}{100} \rfloor = 7$ and $\lfloor r + \frac{k+1}{100} \rfloor = 8$.

We see that if we have a values that return 7 and b values that return 8, then $a+b=73$.

Additionally, because the total sum of all values is 546, we have $7a + 8b = 546$.

Solving the system returns $b=35$, so 35 terms return 8 and the others return 7.

Then, it is notable that the last 35 terms of the series will return 8 (as they are the ones that maximize the value of r), and of course, we have $k+1=91-35+1 =57$. So $k = 56$.

Therefore, the point of change occurs in the step from $\lfloor r + \frac{56}{100} \rfloor$ to $\lfloor r + \frac{57}{100} \rfloor$.

We now use the definition of $\lfloor r + \frac{57}{100} \rfloor = 8$, because if $\lfloor r + \frac{56}{100} \rfloor = 7$, that would imply that $\{r\} = \frac{100-57}{100} = \frac{43}{100}$ (as $\lfloor \frac{56+43}{100} \rfloor = 0$ and $\lfloor \frac{57+43}{100} \rfloor = 1$)

Then, through the definition of $\{r\}$, we see that $r - \lfloor r \rfloor = \{r\}$, and because $\lfloor r \rfloor = 7$ and $\{r\} = \frac{43}{100}$, we get $r = \frac{743}{100} = 7.43$.

$\lfloor 100r \rfloor = \lfloor 100 \cdot 7.43 \rfloor = \lfloor 743 \rfloor =$ $\boxed{743}$

~Pinotation