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Number Theory Level 4

Let a , b , c a,b,c be positive integers such that 1 < a < b < c 1<a<b<c . It is known that only 1 triplet ( a , b , c a,b,c ) satisfies a b c ∣ ( a b − 1 ) ( b c − 1 ) ( c a − 1 ) abc|(ab-1)(bc-1)(ca-1) . Find the value of a + b + c a+b+c .


The answer is 10.

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Victor Loh by Victor Loh , 19, Singapore

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1 solution

a b c ∣ ( a − 1 ) ( b − 1 ) ( c − 1 ) ⇔ a b c ∣ a b + b c + c a − 1 ⇔ K = 1 a + 1 b + 1 c − 1 a b c ∈ Z + abc|(a-1)(b-1)(c-1)\Leftrightarrow abc|ab+bc+ca-1\Leftrightarrow K=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{abc}\in\mathbb{Z^+}

Since 1 < a < b < c ⇒ 2 ≤ a , 3 ≤ b , 4 ≤ c 1<a<b<c\Rightarrow 2\le a, 3\le b, 4\le c , thus K < 2 K<2 so K = 1 K=1 . If a ≥ 3 a\ge 3 , then K < 1 K<1 . Thus it's enough to consider the case a = 2 a=2 , we get ( b − 2 ) ( c − 2 ) = 3 (b-2)(c-2)=3 , ( b , c ) = ( 3 , 5 ) (b,c)=(3,5) .

Hence ( a , b , c ) = ( 2 , 3 , 5 ) ⇒ a + b + c = 10 (a,b,c)=(2,3,5)\Rightarrow a+b+c=\boxed{10}

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How can you say that abc|(a-1)b-1)(c-1) if abc|(ab-1)(bc-1)(ca-1)?

Chaitanya Allu - 6 years, 11 months ago

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Try expanding it and you'll see what happens.

Samuraiwarm Tsunayoshi - 6 years, 11 months ago

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