Divisible by 2000

Number Theory Level 3

Consider two positive integers a a and b b which are such that a b b a a^bb^a is divisible by 2000. What is the least possible value of the product a b ab .


The answer is 20.

Anish Harsha by Anish Harsha , 20, India

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1 solution

Akshat Sharda
Oct 19, 2015

2000 ∣ a b b a ⇒ 2 ∣ a or b and 5 ∣ a or b . Thus, the least possible value of ab = 10. Now possibilities of (a,b) = ( 1 , 10 ) ( 2 , 5 ) ( 5 , 2 ) ( 10 , 1 ) . But in all these cases 2000 doesn’t divide a b b a . The next multiple of 10 is 20. Now 2000 ∣ a b b a in some cases, Answer = 20 2000|a^{b}b^{a}\Rightarrow 2|a \text{ or } b \text{ and } 5|a \text{ or } b. \\ \text{Thus, the least possible value of ab}=10. \\ \text{Now possibilities of (a,b)}=(1,10)(2,5)(5,2)(10,1). \\ \text{But in all these cases } 2000 \text{ doesn't divide } a^{b}b^{a}.\\ \text{The next multiple of }10 \text{ is } 20. \\ \text{Now } 2000|a^{b}b^{a} \text{ in some cases,} \\ \text{Answer}=\boxed{20}

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