1985 AJHSME Problems/Problem 24
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[hide]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
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
Solution 2
Let the numbers, in clockwise order from the very top, be . Notice that . Manipulation will deal . Now we use the answer choices to our advantage. We first check for and find that , which is impossible due to the restrictions in the question. Having eliminated 40, we now move onto and find that this deals , which is possible (for example, let ). Thus, the answer is . -scthecool
See Also
1985 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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