1989 AIME Problems/Problem 1
Solution 1 (Symmetry)
Note that the four numbers to multiply are symmetric with the center at . Multiply the symmetric pairs to get and . .
Solution 2 (Symmetry)
Notice that . Then we can notice that and that . Therefore, . This is because we have that as per the equation .
Solution 3 (Symmetry with Generalization)
More generally, we can prove that one more than the product of four consecutive integers must be a perfect square: At we have ~Novus677 (Fundamental Logic)
Solution 4 (Symmetry with Generalization)
Similar to Solution 1 above, call the consecutive integers to make use of symmetry. Note that itself is not an integer - in this case, . The expression becomes . Distributing each pair of difference of squares first, and then distributing the two resulting quadratics and adding the constant, gives . The inside is a perfect square trinomial, since . It's equal to , which simplifies to . You can plug in the value of from there, or further simplify to , which is easier to compute. Either way, plugging in gives .
Solution 5 (Prime Factorization)
We have Since the alternating sum of the digits is divisible by we conclude that is divisible by
We evaluate the original expression by prime factorization: ~Vrjmath (Fundamental Logic)
Solution 6 (Observation)
The last digit under the radical is , so the square root must either end in or , since means . Additionally, the number must be near , narrowing the reasonable choices to and .
Continuing the logic, the next-to-last digit under the radical is the same as the last digit of , which is . Quick computation shows that ends in , while ends in . Thus, the answer is .
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