Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (mispelling of answer ("anser"))
(→‎Problem)
Line 5: Line 5:
 
</math></div>
 
</math></div>
 
Find <math>\lfloor 100r \rfloor</math>. (For real <math>x^{}_{}</math>, <math>\lfloor x \rfloor</math> is the [[floor function|greatest integer]] less than or equal to <math>x^{}_{}</math>.)
 
Find <math>\lfloor 100r \rfloor</math>. (For real <math>x^{}_{}</math>, <math>\lfloor x \rfloor</math> is the [[floor function|greatest integer]] less than or equal to <math>x^{}_{}</math>.)
 +
 +
==Solution (Hopefully Intuitive)==
 +
Note that the value of <math>r</math> up to the closest multiple of <math>\frac{1}{100}</math> doesn't matter, so assume <math>100r</math> is an integer. By Hermite's Identity, this equation is equivalent to <cmath>\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)</cmath> <cmath>=546.</cmath>
 +
    We can guess that <math>100r</math> is about 600-something. Assuming that <math>608 \le 100r < 682</math>, (so all the floors after 92 are 7 but all the floors before 18 is 6) this equation becomes <cmath>100r - 6 \cdot 18 - 7 \cdot 8 = 546.</cmath> Solving, we get that <cmath>100r = 708,</cmath> which is not in the range that we guessed. Oops. This means that we need for <math>100r</math> to actually go above <math>708</math> since this equation assumes that the sum of the floors is smaller than they actually are once we surpass <math>682</math>.
 +
 +
    We now guess that <math>708 \le 100r < 782</math>. Similar to before, we solve <cmath>100r - 7 \cdot 18 - 8 \cdot 8 = 546</cmath> to get <cmath>100r = 736,</cmath> which is in the range we guessed! So, the answer is <math>736</math>.
 +
 +
~~solasky
  
 
== Solution ==
 
== Solution ==

Revision as of 15:49, 25 July 2023

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 $736$.

~~solasky

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.

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_6&oldid=196073"