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 "2007 iTest Problems/Problem 57"

(Created page with "== Problem == Let <math>T=\text{TNFTPP}</math>. How many positive integers are within <math>T</math> of exactly <math>\lfloor \sqrt T\rfloor</math> perfect squares? (Note: <math>...")
 
(Solution to Problem 57)
 
Line 1: Line 1:
−
== Problem ==
+
''The following problem is from the Ultimate Question of the [[2007 iTest]], where solving this problem required the answer of a previous problem. When the problem is rewritten, the T-value is substituted.''
−
Let <math>T=\text{TNFTPP}</math>. How many positive integers are within <math>T</math> of exactly <math>\lfloor \sqrt T\rfloor</math> perfect squares? (Note: <math>0^2=0</math> is considered a perfect square.)
 
  
−
== Solution ==
+
==Problem==
 +
 
 +
How many positive integers are within <math>810</math> of exactly <math>\lfloor \sqrt{810} \rfloor</math> perfect squares? (Note: <math>0^2=0</math> is considered a perfect square.)
 +
 
 +
==Solution==
 +
 
 +
This problem is essentially asking for how many <math>n</math> are there <math>28</math> perfect squares from <math>n-810</math> to <math>n+810</math>.
 +
 
 +
To find the bounds, note that the difference between consecutive perfect squares are odd numbers.  As it increases, the distance between perfect squares increase.  Let <math>a</math> be the difference between the minimum perfect square and the next perfect square.  Since there are <math>28</math> perfect squares in the range, the last difference is <math>a+54</math>, and the sum of the differences is <math>\tfrac{1}{2} \cdot 28(2a+54)</math>.  This equals <math>1620</math>, so writing an equation and solving for <math>a</math> yields <math>a \approx 30</math>.  The sum of the odd numbers from <math>1</math> to <math>29</math> is <math>225</math>, so we found a place to start.
 +
 
 +
Using the boundries as reference, we can make a table to find values of <math>n</math> and find the number of perfect squares.
 +
 
 +
{|class="wikitable" style="margin-left: auto; margin-right: auto;"
 +
|-
 +
|Value of <math>n</math>
 +
|Value of <math>n-810</math>
 +
|Value of <math>n+810</math>
 +
|Smallest PS in bound
 +
|Largest PS in bound
 +
|Number of PS
 +
|-
 +
|<math>980</math>
 +
|<math>170</math>
 +
|<math>1790</math>
 +
|<math>14^2 = 196</math>
 +
|<math>42^2 = 1764</math>
 +
|<math>29</math>
 +
|-
 +
|<math>1007</math>
 +
|<math>197</math>
 +
|<math>1817</math>
 +
|<math>15^2 = 225</math>
 +
|<math>42^2 = 1764</math>
 +
|<math>28</math>
 +
|-
 +
|<math>1035</math>
 +
|<math>225</math>
 +
|<math>1845</math>
 +
|<math>15^2 = 225</math>
 +
|<math>42^2 = 1764</math>
 +
|<math>28</math>
 +
|-
 +
|<math>1036</math>
 +
|<math>226</math>
 +
|<math>1846</math>
 +
|<math>16^2 = 256</math>
 +
|<math>42^2 = 1764</math>
 +
|<math>27</math>
 +
|-
 +
|<math>1039</math>
 +
|<math>229</math>
 +
|<math>1849</math>
 +
|<math>16^2 = 256</math>
 +
|<math>43^2 = 1849</math>
 +
|<math>28</math>
 +
|-
 +
|<math>1066</math>
 +
|<math>256</math>
 +
|<math>1876</math>
 +
|<math>16^2 = 256</math>
 +
|<math>43^2 = 1849</math>
 +
|<math>28</math>
 +
|}
 +
 +
If <math>n</math> gets below <math>1007</math>, then there will be more perfect squares because <math>12^2 = 144</math> while <math>41^2 = 1681</math>, so more perfect squares would be gained than lost.  Similarly, if <math>n</math> gets above <math>1066</math>, then there will be less perfect squares because <math>18^2 = 324</math> while <math>44^2 = 1936</math>, so more perfect squares would be lost than gained.
 +
 
 +
Based on the table, there are <math>(1035-1007+1)+(1066-1039+1) = \boxed{57}</math> integers that satisfy the criteria.
 +
 
 +
==See Also==
 +
{{iTest box|year=2007|num-b=56|num-a=58}}
 +
 
 +
[[Category:Intermediate Number Theory Problems]]

Latest revision as of 20:25, 26 June 2018

The following problem is from the Ultimate Question of the 2007 iTest, where solving this problem required the answer of a previous problem. When the problem is rewritten, the T-value is substituted.

Problem

How many positive integers are within $810$ of exactly $\lfloor \sqrt{810} \rfloor$ perfect squares? (Note: $0^2=0$ is considered a perfect square.)

Solution

This problem is essentially asking for how many $n$ are there $28$ perfect squares from $n-810$ to $n+810$.

To find the bounds, note that the difference between consecutive perfect squares are odd numbers. As it increases, the distance between perfect squares increase. Let $a$ be the difference between the minimum perfect square and the next perfect square. Since there are $28$ perfect squares in the range, the last difference is $a+54$, and the sum of the differences is $\tfrac{1}{2} \cdot 28(2a+54)$. This equals $1620$, so writing an equation and solving for $a$ yields $a \approx 30$. The sum of the odd numbers from $1$ to $29$ is $225$, so we found a place to start.

Using the boundries as reference, we can make a table to find values of $n$ and find the number of perfect squares.

Value of $n$ Value of $n-810$ Value of $n+810$ Smallest PS in bound Largest PS in bound Number of PS
$980$ $170$ $1790$ $14^2 = 196$ $42^2 = 1764$ $29$
$1007$ $197$ $1817$ $15^2 = 225$ $42^2 = 1764$ $28$
$1035$ $225$ $1845$ $15^2 = 225$ $42^2 = 1764$ $28$
$1036$ $226$ $1846$ $16^2 = 256$ $42^2 = 1764$ $27$
$1039$ $229$ $1849$ $16^2 = 256$ $43^2 = 1849$ $28$
$1066$ $256$ $1876$ $16^2 = 256$ $43^2 = 1849$ $28$

If $n$ gets below $1007$, then there will be more perfect squares because $12^2 = 144$ while $41^2 = 1681$, so more perfect squares would be gained than lost. Similarly, if $n$ gets above $1066$, then there will be less perfect squares because $18^2 = 324$ while $44^2 = 1936$, so more perfect squares would be lost than gained.

Based on the table, there are $(1035-1007+1)+(1066-1039+1) = \boxed{57}$ integers that satisfy the criteria.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 56 		Followed by:
Problem 58
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_iTest_Problems/Problem_57&oldid=95639"