1985 AJHSME Problems/Problem 24
Contents
Problem
In a magic triangle, each of the six whole numbers is placed in one of the circles so that the sum, , of the three numbers on each side of the triangle is the same. The largest possible value for is
Solution
Let the number in the top circle be and then , , , , and , going in clockwise order. Then, we have
Adding these equations together, we get
where the last step comes from the fact that since , , , , , and are the numbers in some order, their sum is
The left hand side is divisible by and is divisible by , so must be divisible by . The largest possible value of is then , and the corresponding value of is , which is choice .
It turns out this sum is attainable if you let
Solution 2
To make the sum the greatest, put the three largest numbers and in the corners. Then, balance the sides by putting the least integer between the greatest sum and. Then put the next least integer between the next greatest sum (). Fill in the last integer and you can see that the sum of any three numbers on a side is (for example) . -by goldenn
See Also
1985 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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