Online Course Discussion Forum

Please help with winter camp Number theory p19-21

 
 
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Please help with winter camp Number theory p19-21
by Iris Xu - Wednesday, December 30, 2020, 12:46 AM
 

not sure about where to start. any help would be appreciated!

 
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Re: Please help with winter camp Number theory p19-21
by Iris Xu - Wednesday, December 30, 2020, 10:46 PM
 

Update: I now need help on number theory blue p15,  p19 and p20. thanks!

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Re: Please help with winter camp Number theory p19-21
by Charles Zhang - Thursday, December 31, 2020, 9:57 AM
 

So far, I have only finished 15.


Hint: Prime Factorize 2013 and remember 7!/6! = 7 (how can you use this for 2013 similarly?)


Hint for after you prime factorize: you must MINIMIZE a_0 + b_0.  61 is prime, 59 as well.  How can you build an extra "60" term in the denominator?

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Re: Please help with winter camp Number theory p19-21
by Areteem Professor - Wednesday, January 20, 2021, 2:50 PM
 

Number Theory 19: Note the problem is equivalent to "What is the smallest value of $k$ for which $k(k+1)(2k+1)$ is a multiple of $1200$. Thus, $3 \mid k(k+1)(2k+1)$, $2^4 \mid k(k+1)(2k+1)$, and $5^2 \mid k(k+1)(2k+1)$.  Use this to narrow down possible choices for $k$.

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Re: Please help with winter camp Number theory p19-21
by Areteem Professor - Wednesday, January 20, 2021, 2:51 PM
 

Number Theory 20: Since $a = 2^n \cdot 3^m$, $a^6 = 2^{6n} \cdot 3^{6m}$, and $$\begin{aligned} 6^a &= (2 \cdot 3)^a \\ &= (2 \cdot 3)^{2^n \cdot 3^m} \\ &= 2^{2^n \cdot 3^m} \cdot 3^{2^n \cdot 3^m} \end{aligned}.$$ Look at the exponents of both. What do you need to ask of $n$ and $m$ so that $a^6 \mid 6^a$?