Let $n$ be the smallest positive integer such that $mn$ is a perfect $k$th power of an integer for some $k \ge 2$, where $m=2^{1980} \cdot 3^{384} \cdot 5^{1694} \cdot 7^{343}$. What is $n+k$?

Guest Jun 27, 2017

2+0 Answers


Can't understand your question with all those $$ signs !!.

Guest Jun 27, 2017

\(\text{The first solution that comes to mind is choosing }n=7\\ \text{and }k=2\text{. This allows the product }mn\text{ to consist of}\\ \text{even powers which results in a natural number when }\\ \text{taking the second power root. Now we need to check if}\\ \text{this solution is indeed the smallest possible value for }n.\\ \text{We know that a smaller solution can be acquired only}\\ \text{if we don't need to add an extra factor 7. This requires}\\ \text{343 being divisible by }k\text{. We know }343=7^3\text{ so }k\text{ must}\\ \text{equal some power of 7 for this to be the case. 1980 and}\\ \text{384 are not divisible by 7, 1694 however is. To correct for}\\ \text{this }n\text{ needs to equal 6 (because 1981 and 385 are divisible}\\ \text{7). As 6 is smaller than 7 are solution becomes: }n=6,\\ k=7;n+k=13.\\ \text{If this explanation is still a bit hazy, allow me to show it in}\\ \text{formula form:}\\ \)

\(n=2^\alpha\cdot3^\beta\cdot5^\gamma\cdot7^\delta\Rightarrow\\ mn=2^{1980+\alpha}\cdot3^{383+\beta}\cdot5^{1694+\gamma}\cdot7^{343+\delta}\Rightarrow\\ 2^{\frac{1980+\alpha}k}\cdot3^{\frac{383+\beta}k}\cdot5^{\frac{1694+\gamma}k}\cdot7^{\frac{343+\delta}k}\in\mathbb{N}\Rightarrow\\ \frac{1980+\alpha}k,\frac{383+\beta}k,\frac{1694+\gamma}k,\frac{343+\delta}k\in\mathbb{N}.\)

Honga  Jun 27, 2017

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