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Please help with winter camp Number theory p19-21

 
 
XuIris的头像
Please help with winter camp Number theory p19-21
XuIris - 2020年12月30日 Wednesday 00:46
 

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

 
XuIris的头像
Re: Please help with winter camp Number theory p19-21
XuIris - 2020年12月30日 Wednesday 22:46
 

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

ZhangCharles的头像
Re: Please help with winter camp Number theory p19-21
ZhangCharles - 2020年12月31日 Thursday 09:57
 

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?

ProfessorAreteem的头像
Re: Please help with winter camp Number theory p19-21
ProfessorAreteem - 2021年01月20日 Wednesday 14:50
 

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 $k$.

ProfessorAreteem的头像
Re: Please help with winter camp Number theory p19-21
ProfessorAreteem - 2021年01月20日 Wednesday 14:51
 

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 $a^6 \mid 6^a$?