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

6 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details