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| − | ==Problem==
| + | #redirect [[2020 AMC 10B Problems/Problem 24]] |
| − | How many positive integers <math>n</math> satisfy<cmath>\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?</cmath>(Recall that <math>\lfloor x\rfloor</math> is the greatest integer not exceeding <math>x</math>.)
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| − | <math>\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32</math>
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| − | ==Solution==
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| − | *Not a reliable or in-depth solution (for the guess and check students)
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| − | We can first consider the equation without a floor function:
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| − | <cmath>\dfrac{n+1000}{70} = \sqrt{n} </cmath>
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| − | Multiplying both sides by 70 and then squaring:
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| − | <cmath>n^2 + 2000n + 1000000 = 4900n</cmath>
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| − | Moving all terms to the left:
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| − | <cmath>n^2 - 2900n + 1000000 = 0</cmath>
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| − | Now we can use wishful thinking to determine the factors:
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| − | <cmath>(n-400)(n-2500) = 0</cmath>
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| − | This means that for <math>n = 400</math> and <math>n = 2500</math>, the equation will hold without the floor function.
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| − | Now we can simply check the multiples of 70 around 400 and 2500 in the original equation:
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| − | For <math>n = 330</math>, left hand side = <math>19</math> but <math>18^2 < 330 < 19^2</math> so right hand side = <math>18</math>
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| − | For <math>n = 400</math>, left hand side = <math>20</math> and right hand side = <math>20</math>
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| − | For <math>n = 470</math>, left hand side = <math>21</math> and right hand side = <math>21</math>
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| − | For <math>n = 540</math>, left hand side = <math>22</math> but <math>540 > 23^2</math> so right hand side = <math>23</math>
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| − | Now we move to <math>n = 2500</math>
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| − | For <math>n = 2430</math>, left hand side = <math>49</math> and <math>49^2 < 2430 < 50^2</math> so right hand side = <math>49</math>
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| − | For <math>n = 2360</math>, left hand side = <math>48</math> and <math>48^2 < 2360 < 49^2</math> so right hand side = <math>48</math>
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| − | For <math>n = 2290</math>, left hand side = <math>47</math> and <math>47^2 < 2360 < 48^2</math> so right hand side = <math>47</math>
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| − | For <math>n = 2220</math>, left hand side = <math>46</math> but <math>47^2 < 2220</math> so right hand side = <math>47</math>
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| − | For <math>n = 2500</math>, left hand side = <math>50</math> and right hand side = <math>50</math>
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| − | For <math>n = 2570</math>, left hand side = <math>51</math> but <math>2570 < 51^2</math> so right hand side = <math>50</math>
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| − | Therefore we have 6 total solutions, <math>n = 400, 470, 2290, 2360, 2430, 2500 = \boxed{\textbf{(C) 6}}</math>
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| − | ==Solution 2==
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| − | This is my first solution here, so please forgive me for any errors.
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| − | We are given that <cmath>\frac{n+1000}{70}=\lfloor\sqrt{n}\rfloor</cmath>
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| − | <math>\lfloor\sqrt{n}\rfloor</math> must be an integer, which means that <math>n+1000</math> is divisible by <math>70</math> As <math>1000\equiv 20\pmod{70}</math>, this means that <math>n\equiv 50\pmod{70}</math>, so we can write <math>n=70k+50</math> for an integer <math>k</math>.
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| − | Therefore, <cmath>\frac{n+1000}{70}=\frac{70k+1050}{70}=k+15=\lfloor\sqrt{70k+50}\rfloor</cmath>
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| − | Also, we can say that <math>\sqrt{70k+50}-1\leq k+15</math> and <math>k+15\leq\sqrt{70k+50}</math>
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| − | Squaring the second inequality, we get <math>k^{2}+30k+225\leq70k+50\implies k^{2}-40k+175\leq 0\implies (k-5)(k-35)\leq0\implies 5\leq k\leq 35</math>.
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| − | Similarly solving the first inequality gives us <math>k\leq 19-\sqrt{155}</math> or <math>k\geq 19+\sqrt{155}</math>
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| − | <math>\sqrt{155}</math> is slightly greater than <math>12</math>, so instead, we can say <math>k\leq 6</math> or <math>k\geq 32</math>.
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| − | Combining this with <math>5\leq k\leq 35</math>, we get <math>k=5,6,32,33,34,35</math> are all solutions for <math>k</math> that give a valid solution for <math>n</math>, meaning that our answer is <math>\boxed{6}</math>.
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| − | ==See Also==
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| − | {{AMC12 box|year=2020|ab=B|num-b=20|num-a=22}}
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| − | {{MAA Notice}}
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