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what is the largest value of n for which n^2 is a factor of 6^101+10^101?

 Aug 13, 2020
 #1
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1 - First factor [6^101  +  10^101]

 

2 - You should get:   

 2^104 × 2 217557 × 265 896803 424061 × 1568 372427 881573 587879 × 5331 425490 815171 853272 879027

 

3 - Multiply  2^104 =20,282,409,603,651,670,423,947,251,286,016

 

4 - Take the square root of the result in 3 above:

 

5 - You should get: 4,503,599,627,370,496 - This is the largest n. So that:

 

6 - 4,503,599,627,370,496^2 = Result in 3 above which is a factor of 6^101 + 10^101.

 Aug 13, 2020
edited by Guest  Aug 13, 2020
 #2
avatar+31340 
+2

Here's another way of doing this for those who don't have a handy facility for factorising large numbers:

 

 Aug 13, 2020
 #3
avatar+25598 
+3

what is the largest value of \(n\) for which \(n^2\) is a factor of \(6^{101}+10^{101}\)?

 

Formula:

For an odd positive integer \(n\)
\(a^n+b^n=\left(a+b \right)\left(a^{n-1}-a^{n-2}b+\cdots -ab^{n-2}+b^{n-1}\right) \)

Source "What is the formula for \(a^n + b^n\)?" see: https://www.quora.com/What-is-the-formula-for-a-n-b-n

 

\(\begin{array}{|rcll|} \hline \mathbf{a^n+b^n} &=& \mathbf{\left(a+b \right)\left(a^{n-1}-a^{n-2}b+\cdots -ab^{n-2}+b^{n-1}\right)} \\ && \boxed{n=101,\ a=6,\ b=10} \\\\ 6^{101}+10^{101} &=& (6+10)\left(6^{101-1}-6^{101-2}*10+\cdots -6*10^{101-2}+10^{101-1}\right) \\\\ 6^{101}+10^{101} &=& 16\left(6^{100}-6^{99}*10+\cdots -6*10^{99}+10^{100}\right) \\\\ 6^{101}+10^{101} &=& 2^4\left(6^{100}-6^{99}*10+6^{98}*10^{2}-6^{97}*10^{3}+ \\\\ \cdots -6^{3}*10^{97}+6^{2}*10^{98}-6*10^{99}+10^{100}\right) \\\\ 6^{101}+10^{101} &=& 2^4\left(2^{100}3^{100}-2^{99}3^{99}*2*5+2^{98}3^{98}*2^{2}*5^{2}-2^{97}3^{97}*2^{3}*5^{3}+ \\\\ \cdots -2^{3}3^{3}*2^{97}5^{97}+2^{2}3^{2}*2^{98}5^{98}-2*3*2^{99}5^{99}+2^{100}5^{100}\right) \\\\ 6^{101}+10^{101} &=& 2^4\left(2^{100}3^{100}-2^{100}3^{99}*5+2^{100}3^{98}*5^{2}-2^{100}3^{97}*5^{3}+ \\\\ \cdots -2^{100}3^{3}*5^{97}+2^{100}3^{2}*5^{98}-2^{100}*3*5^{99}+2^{100}5^{100}\right) \\\\ 6^{101}+10^{101} &=& 2^42^{100}\left(3^{100}-3^{99}*5+3^{98}*5^{2}-3^{97}*5^{3}+ \\\\ \cdots -3^{3}*5^{97}+3^{2}*5^{98}-3*5^{99}+5^{100}\right) \\\\ 6^{101}+10^{101} &=& \mathbf{{\color{red}2^{104}}}\left(3^{100}-3^{99}*5+3^{98}*5^{2}-3^{97}*5^{3}+ \\\\ \cdots -3^{3}*5^{97}+3^{2}*5^{98}-3*5^{99}+5^{100}\right) \\ \hline n^2 &=& \mathbf{{\color{red}2^{104}}} \\ n &=& 2^{\frac{104}{2}} \\ \mathbf{n} &=& \mathbf{2^{52}} \\ \hline \end{array}\)

 

without latex:

 

laugh

 Aug 14, 2020
edited by heureka  Aug 14, 2020

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