Difference between revisions of "1991 AIME Problems/Problem 6"

Revision as of 10:23, 1 October 2020

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

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 anser 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.

@@ Line 12: / Line 12: @@
 Recall by Hermite's Identity that <math>\lfloor x\rfloor +\lfloor x+\frac{1}{n}\rfloor +...+\lfloor x+\frac{n-1}{n}\rfloor = \lfloor nx\rfloor</math> for positive integers <math>n</math>, and real <math>x</math>. 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, <math>\lfloor r\rfloor \le ...\le \lfloor r+\frac{18}{100}\rfloor \le 7</math> and <math>\lfloor r+\frac{92}{100}\rfloor \ge ...\ge \lfloor r+1\rfloor \ge 8</math>. We can see that <math>\lfloor r\rfloor +1=\lfloor r+1\rfloor</math>. Because <math>\lfloor r\rfloor</math> is at most 7, and <math>\lfloor r+1\rfloor</math> is at least 8, we can clearly see their values are <math>7</math> and <math>8</math> respectively.
 So,  <math>\lfloor r\rfloor = ... = \lfloor r+\frac{18}{100}\rfloor = 7</math>, and <math>\lfloor r+\frac{92}{100}\rfloor = ...= \lfloor r+1\rfloor = 8</math>. Since there are 19 terms in the former equation and 8 terms in the latter, our anser is <math>\lfloor nx\rfloor = 546+19\cdot 7+8\cdot 8=\boxed{743}</math>
+=== Note ===
+In the contest, you would just observe this mentally, and then calculate <math>546+19\cdot 7+8\cdot 8= 743</math>, hence the speed at which one can carry out this solution.
 == See also ==

1991 AIME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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