1985 AJHSME Problems/Problem 24
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 1
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
Video Solution
~savannahsolver
See Also
1985 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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