2019 AMC 12A Problems/Problem 15
Positive real numbers and have the property that and all four terms on the left are positive integers, where log denotes the base 10 logarithm. What is ?
Since is a positive integer, we get for some integer ; since is a positive integer, we get . Thus ; similarly . Substituting, we get , i.e. . It follows that . The values of for are
Two of those values must add up to and we see that , so and , and our answer is .
Since all four terms on the left are positive integers, from , we know that both has to be a perfect square and has to be a power of ten. The same applies to for the same reason. Setting and to and , where and are the perfect squares, . By listing all the perfect squares up to (as is larger than the largest possible sum of and of from answer choice ), two of those perfect squares must add up to one of the possible sums of and given from the answer choices (, , , , or ).
Only a few possible sums are seen: , , , , and . By testing each of these (seeing whether ), only the pair and work. Therefore, and are and , and our answer is .
Given that and are both integers, and must be in the form and , respectively for some positive integers and . Note that . By substituting for a and b, the equation becomes . After multiplying the equation by 2 and completing the square with respect to and , the equation becomes . Testing squares of positive integers that add to , is the only option. Without loss of generality, let and . Plugging in and to solve for and gives us and . Therefore, .
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