CEMC Fermat

Algebra Difficulty 2.6 Junior Find the answer

Positive integers $a$ and $b$ satisfy $a b=2010$ . If $a>b$ , what is the smallest possible value of $a-b$ ?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Note that $2010=10(201)=2(5)(3)(67)$ and that 67 is prime. Therefore, the positive divisors of 2010 are $1,2,3,5,6,10,15,30,67,134,201,335,402,670$ , $1005,2010$ . Thus, the possible pairs $(a, b)$ with $a b=2010$ and $a>b$ are $(2010,1),(1005,2),(670,3)$ , $(402,5),(335,6),(201,10),(134,15),(67,30)$ . Of these pairs, the one with the smallest possible value of $a-b$ is $(a, b)=(67,30)$ , which gives $a-b=37$ .

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