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

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== Solution 2 (Faster) ==
 
== Solution 2 (Faster) ==
−
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,...,\lfloor r+\frac{18}{100}\rfloor \le 7</math> and <math>\lfloor r+\frac{92}{100},...,\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 nx\rfloor</math> is <math>546+19\cdot 7+8\cdot 8=\boxed{743}</math>
+
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,...,\lfloor r+\frac{18}{100}\rfloor \le 7</math> and <math>\lfloor r+\frac{92}{100}\rfloor,...,\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 nx\rfloor</math> is <math>546+19\cdot 7+8\cdot 8=\boxed{743}</math>
  
 
== See also ==
 
== See also ==

Revision as of 10:18, 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, $\lfloor r\rfloor,...,\lfloor r+\frac{18}{100}\rfloor \le 7$ and $\lfloor r+\frac{92}{100}\rfloor,...,\lfloor r+1\rfloor \ge 8$. 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, $lfloor nx\rfloor$ is $546+19\cdot 7+8\cdot 8=\boxed{743}$

See also

1991 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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