 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2004/c/c1/a/ii/solution

Write

a = \prod_{p} p^{r_{p}}

and

b = \prod_{p} p^{s_{p}}

min (r_{p}, s_{p})

max (r_{p}, s_{p})

. Since their sum is

r_{p} + s_{p}

for each prime,

g cd (a, b) lcm (a, b) = p \prod p^{r_{p} + s_{p}} = ab .

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