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Finite Math, 2.29 Math Challenge I-C

 
 
WangChristine的头像
Finite Math, 2.29 Math Challenge I-C
WangChristine - 2022年03月19日 Saturday 11:34
 
In this question, I'm a little stuck, so is it  okay for you to send me the solution for this problem? I watched the playback for the live discussion, but I don't think the teacher got to talk about this question. Thanks a lot! 
 
LensmireJohn的头像
Re: Finite Math, 2.29 Math Challenge I-C
LensmireJohn - 2022年03月21日 Monday 13:11
 

Remember that now that the HW is submitted, you can review the answers and main ideas for the solutions on the course page too (by reviewing the quiz).

Let me give some hints to expand on and motivate what's happening in the solution. The key thing is looking carefully at the information given and what it tells us about A, B, and C (plus their prime factorizations).

  • $\gcd(A,B) = 12$. This means that $A$ and $B$ are multiples of $12 = 2^2\cdot 3$.
  • $\text{lcm}(A,B) = 396$. This means that $A$ and $B$ are factors of $396 = 2^2\cdot 3^2\cdot 11$.

Pausing here, note we actually know a lot about $A$ and $B$ and their prime factorizations. In fact, $$A = 2^2\cdot 3^{1\text{ or } 2}\cdot 11^{0\text{ or }1}$$ and, similarly, $$B = 2^2\cdot 3^{1\text{ or } 2}\cdot 11^{0\text{ or }1}.$$ Further (and this is important to understand), if $A$ has a $3^2$ term then $B$ must have a $3^1$ term, else their GCD would be larger than $12$ or their LCM would be less than $396$. This is true in reverse, and something similar happens for the exponents of $11$ as well. Hence, there are actually not too many cases to consider.

Lastly, we haven't even mentioned that we also need $\gcd(B,C) = 33$. This implies that $B$ (and $C$) are multiples of $33 = 3\cdot 11$, which actually narrows down things even more.

Hope this and the solution sketch when you review the HW helps!