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Difference between revisions of "2020 AMC 10B Problems/Problem 24"

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==Solution==
 
==Solution==
 +
(Quick solution if you’re in a hurry)
 +
(Could someone fix this with LaTeX?)
 +
First notice that (n+1000)/70 must be an integer. This means that n is congruent to 50 (mod 70). Testing the first few values of n (adding 70 to n each time and noticing the left side increases by 1 each time) yields n=20 and n=21. After this, the right side becomes greater than the left. Since the left side is linear and the right side is roughly a radical, the graphs must intersect in another place. Using binary search can give a good estimate of the other cases, being n=47, n=48, n=49, and n=50.
  
 
==See Also==  
 
==See Also==  

Revision as of 02:29, 8 February 2020

Problem

How many positive integers $n$ satisfy\[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\](Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$.)

$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$

Solution

(Quick solution if you’re in a hurry) (Could someone fix this with LaTeX?) First notice that (n+1000)/70 must be an integer. This means that n is congruent to 50 (mod 70). Testing the first few values of n (adding 70 to n each time and noticing the left side increases by 1 each time) yields n=20 and n=21. After this, the right side becomes greater than the left. Since the left side is linear and the right side is roughly a radical, the graphs must intersect in another place. Using binary search can give a good estimate of the other cases, being n=47, n=48, n=49, and n=50.

See Also

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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
All AMC 10 Problems and Solutions

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