# 1991 AIME Problems/Problem 6

## Problem

Suppose $r^{}_{}$ is a real number for which $\left\lfloor r + \frac{19}{100} \right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100} \right\rfloor + \cdots + \left\lfloor r + \frac{91}{100} \right\rfloor = 546.$

Find $\lfloor 100r \rfloor$. (For real $x^{}_{}$, $\lfloor x \rfloor$ is the greatest integer less than or equal to $x^{}_{}$.)

## Solution (Hopefully Intuitive)

Note that the value of $r$ up to the closest multiple of $\frac{1}{100}$ doesn't matter, so assume $100r$ is an integer. By Hermite's Identity, this equation is equivalent to $$\left\lfloor 100r \right\rfloor - \left(\left\lfloor r + \frac{1}{100}\right\rfloor + \cdots + \left\lfloor r + \frac{18}{100}\right\rfloor + \left\lfloor r + \frac{92}{100}\right\rfloor + \cdots + \left\lfloor r + \frac{99}{100}\right\rfloor \right)$$ $$=546.$$

   We can guess that $100r$ is about 600-something. Assuming that $608 \le 100r < 682$, (so all the floors after 92 are 7 but all the floors before 18 is 6) this equation becomes $$100r - 6 \cdot 18 - 7 \cdot 8 = 546.$$ Solving, we get that $$100r = 708,$$ which is not in the range that we guessed. Oops. This means that we need for $100r$ to actually go above $708$ since this equation assumes that the sum of the floors is smaller than they actually are once we surpass $682$.

   We now guess that $708 \le 100r < 782$. Similar to before, we solve $$100r - 7 \cdot 18 - 8 \cdot 8 = 546$$ to get $$100r = 736,$$ which is in the range we guessed! So, the answer is $736$.


## Solution

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 $\lfloor r+\frac{92}{100}\rfloor \ge ...\ge \lfloor r+1\rfloor \ge 8$. 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 $\lfloor nx\rfloor = 546+19\cdot 7+8\cdot 8=\boxed{743}$

### 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.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 