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Difference between revisions of "2006 SMT/General Problems/Problem 10"

(Created page with "==Solution== First of all, lets note that the sum of all positive integers from <math>1</math> to <math>n</math> inclusive is <math>\frac{n(n+1)}{2}</math>. The sum of all nu...")
 
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==Problem==
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What is the square root of the sum of the first <math> 2006 </math> positive odd integers?
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==Solution==
 
==Solution==
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The sum of the first n positive odd integers is <math>n^2</math>. This comes from the fact that <math>(n+1)^2-n^2 = 2n+1</math> (Taking a sum of this equation beginning with <math>n = 0</math> will yield the desired result as the LHS will telescope). Therefore, the sum of the first 2006 positive odd integers is <math>2006^2</math>. The answer we are looking for is <math>\sqrt{2006^2} = \boxed{2006}</math>
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First of all, lets note that the sum of all positive integers from <math>1</math> to <math>n</math> inclusive is <math>\frac{n(n+1)}{2}</math>. The sum of all numbers from <math>1</math> to <math>2006</math> is then:
 
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<cmath>\frac{2006(2007)}{2} = (1003)(2007)</cmath>
 
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Finding the prime factorization of the product, we see that:
 
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<cmath>(1003)(2007)=17 \cdot 59 \cdot 3^2 \cdot 223</cmath>
 
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Taking the square root, the answer is:
 
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<cmath>\sqrt{(1003)(2007)}=\sqrt{17 \cdot 59 \cdot 3^2 \cdot 223} = 3\sqrt{17 \cdot 59 \cdot 223} = \boxed{3\sqrt{223669}}</cmath>
 

Latest revision as of 17:36, 14 January 2020

Problem

What is the square root of the sum of the first $2006$ positive odd integers?

Solution

The sum of the first n positive odd integers is $n^2$. This comes from the fact that $(n+1)^2-n^2 = 2n+1$ (Taking a sum of this equation beginning with $n = 0$ will yield the desired result as the LHS will telescope). Therefore, the sum of the first 2006 positive odd integers is $2006^2$. The answer we are looking for is $\sqrt{2006^2} = \boxed{2006}$

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