Difference between revisions of "2007 AIME I Problems/Problem 11"
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== Problem == | == Problem == | ||
− | For each [[positive]] [[integer]] <math>p</math>, let <math>b(p)</math> denote the unique positive integer <math>k</math> such that <math>|k-\sqrt{p}| < \frac{1}{2}</math>. For example, <math>b(6) = 2</math> and <math>b(23) = 5</math>. If <math>S = \ | + | For each [[positive]] [[integer]] <math>p</math>, let <math>b(p)</math> denote the unique positive integer <math>k</math> such that <math>|k-\sqrt{p}| < \frac{1}{2}</math>. For example, <math>b(6) = 2</math> and <math>b(23) = 5</math>. If <math>S = \sum_{p=1}^{2007} b(p),</math> find the [[remainder]] when <math>S</math> is divided by 1000. |
− | == Solution == | + | == Solution 1 == |
− | <math>(k- \frac 12)^2=k^2-k+\frac 14</math> and<math>(k+ \frac 12)^2=k^2+k+ \frac 14</math> Therefore <math>b(p)=k</math> if and only if <math>p</math> is in this range, | + | <math>\left(k- \frac 12\right)^2=k^2-k+\frac 14</math> and <math>\left(k+ \frac 12\right)^2=k^2+k+ \frac 14</math>. Therefore <math>b(p)=k</math> if and only if <math>p</math> is in this range, or <math>k^2-k<p\leq k^2+k</math>. There are <math>2k</math> numbers in this range, so the sum of <math>b(p)</math> over this range is <math>(2k)k=2k^2</math>. <math>44<\sqrt{2007}<45</math>, so all numbers <math>1</math> to <math>44</math> have their full range. Summing this up with the formula for the sum of the first <math>n</math> squares (<math>\frac{n(n+1)(2n+1)}{6}</math>), we get <math>\sum_{k=1}^{44}2k^2=2\frac{44(44+1)(2*44+1)}{6}=58740</math>. We need only consider the <math>740</math> because we are working with modulo <math>1000</math>. |
− | Now consider the range of numbers such that <math> | + | Now consider the range of numbers such that <math>b(p)=45</math>. These numbers are <math>\left\lceil\frac{44^2 + 45^2}{2}\right\rceil = 1981</math> to <math>2007</math>. There are <math>2007 - 1981 + 1 = 27</math> (1 to be inclusive) of them. <math>27*45=1215</math>, and <math>215+740= |
+ | \boxed{955}</math>, the answer. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>p</math> be in the range of <math>a^2 \le p < (a+1)^2</math>. Then, we need to find the point where the value of <math>b(p)</math> flips from <math>k</math> to <math>k+1</math>. This will happen when <math>p</math> exceeds <math>(a+\frac{1}{2})^2</math> or <math>a(a+1)+\frac{1}{4}</math>. Thus, if <math>a^2 \le p \le a(a+1)</math> then <math>b(p)=a</math>. For <math>a(a+1) < p < (a+1)^2</math>, then <math>b(p)=a+1</math>. There are <math>a+1</math> terms in the first set of <math>p</math>, and <math>a</math> terms in the second set. Thus, the sum of <math>b(p)</math> from <math>a^2 \le p <(a+1)^2</math> is <math>2a(a+1)</math> or <math>4\cdot\binom{a+1}{2}</math>. For the time being, consider that <math>S = \sum_{p=1}^{44^2-1} b(p)</math>. Then, the sum of the values of <math>b(p)</math> is <math>4\binom{2}{2}+4\binom{3}{2}+\cdots +4\binom{44}{2}=4\left(\binom{2}{2}+\binom{3}{2}+\cdots +\binom{44}{2}\right)</math>. We can collapse this to <math>4\binom{45}{3}=56760</math>. Now, we have to consider <math>p</math> from <math>44^2 \le p < 2007</math>. Considering <math>p</math> from just <math>44^2 \le p \le 1980</math>, we see that all of these values have <math>b(p)=44</math>. Because there are <math>45</math> values of <math>p</math> in that range, the sum of <math>b(p)</math> in that range is <math>45\cdot44=1980</math>. Adding this to <math>56760</math> we get <math>58740</math> or <math>740</math> mod <math>1000</math>. Now, take the range <math>1980 < p \le 2007</math>. There are <math>27</math> values of <math>p</math> in this range, and each has <math>b(p)=45</math>. Thus, that contributes <math>27*45=1215</math> or <math>215</math> to the sum. Finally, adding <math>740</math> and <math>215</math> we get <math>740+215=\boxed{955}</math>. | ||
+ | |||
+ | ~firebolt360 | ||
== See also == | == See also == | ||
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[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 19:02, 30 December 2023
Contents
Problem
For each positive integer , let denote the unique positive integer such that . For example, and . If find the remainder when is divided by 1000.
Solution 1
and . Therefore if and only if is in this range, or . There are numbers in this range, so the sum of over this range is . , so all numbers to have their full range. Summing this up with the formula for the sum of the first squares (), we get . We need only consider the because we are working with modulo .
Now consider the range of numbers such that . These numbers are to . There are (1 to be inclusive) of them. , and , the answer.
Solution 2
Let be in the range of . Then, we need to find the point where the value of flips from to . This will happen when exceeds or . Thus, if then . For , then . There are terms in the first set of , and terms in the second set. Thus, the sum of from is or . For the time being, consider that . Then, the sum of the values of is . We can collapse this to . Now, we have to consider from . Considering from just , we see that all of these values have . Because there are values of in that range, the sum of in that range is . Adding this to we get or mod . Now, take the range . There are values of in this range, and each has . Thus, that contributes or to the sum. Finally, adding and we get .
~firebolt360
See also
2007 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.