What A Way To Factorize It!

Number Theory Level 2

$\large 2000 \mid a^b \times b^a$

Let $a$ and $b$ be two positive integers such that ${ a }^{ b }\times { b }^{ a }$ is divisible by $2000$ . Find the least value of the product of $a$ and $b$ .

The answer is 20.

1 solution

Akshat Sharda
Mar 3, 2016

$2000|a^b b^a \Rightarrow \begin{cases} 2| a \text{ or }b \\ 5| a \text{ or } b \end{cases} \\ \therefore 10|ab \\ \text{If }ab \text{ then the possibilities are,} \\ (a,b)=(1,10),(2,5),(5,2),(10,1) \\ \text{But in all the cases, it is easy to see that }2000\text{ doesn't divide }ab. \\ \text{Next multiple of }10\text{ is }20, \\ \therefore (a,b)=(4,5) \text{ and it also satisfies our conditions, } \\ 2000 \text{ divides }4^5 5^4 \Rightarrow \therefore ab=\boxed{20}$

I used the fact that $2000=2^{4}*5^{3}$ . Since 3>2, then 3 cannot be the power of 5. Using $2^{2}=4$ as the power of 5 leaves $5^{4}*4^{5}$ . I don't know if that's valid approach to determining the smallest multiple of 2000 but that's how I found it.

Jerry McKenzie - 4 years, 1 month ago

My method was not as rigorous as this - I calculated that if a = 4 and b = 5 (or vice versa) the equation is satisfied, and then checked values lower than these to see if they would work, and when I had satisfied myself that they would not I gave my answer.

Thomas Sutcliffe - 3 years, 2 months ago

What A Way To Factorize It!

The answer is 20.

1 solution

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