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Math challenge I-A homework

 
 
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Re: Math challenge I-A homework
by Areteem Professor - Wednesday, April 1, 2020, 11:51 AM
 

There's two approaches you can follow for this problem:

1) Let $m = p_1^{e_1} \cdot p_2^{e_2} \cdots p_k^{e_k}$ and $n = p_1^{f_1} \cdot p_2^{f_2} \cdots p_k^{f_k}$ be two positive integers (and their prime factorizations). Then $$\text{lcm} (m,n) = p_1^{\max(e_1,f_1)}\cdot p_2^{\max(e_2,f_2)}\cdots p_k^{\max(e_k,f_k)}$$ and $$\gcd (m,n) = p_1^{\min(e_1,f_1)}\cdot p_2^{\min(e_2,f_2)}\cdots p_k^{\min(e_k,f_k)}.$$ For example, $48 = 2^4 \cdot 3 \cdot 5^0$ and $180 = 2^2 \cdot 3^2 \cdot 5$, so $$\text{lcm}(48,180) = 2^4 \cdot 3^2\cdot 5 = 720$$ and $$\gcd(48,180) = 2^2\cdot 3 \cdot 5^0 = 12.$$ So you can look at the prime factorizations of the given numbers to figure out the prime factorization of the missing number.

2) From what we have above, it is easy to check that for any two integers $m$ and $n$ it is always true that $$m\cdot n = \text{lcm}(m,n) \cdot \gcd(m,n).$$