Difference between revisions of "1985 AJHSME Problems/Problem 24"
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<math>\text{(A)}\ 36 \qquad \text{(B)}\ 37 \qquad \text{(C)}\ 38 \qquad \text{(D)}\ 39 \qquad \text{(E)}\ 40</math> | <math>\text{(A)}\ 36 \qquad \text{(B)}\ 37 \qquad \text{(C)}\ 38 \qquad \text{(D)}\ 39 \qquad \text{(E)}\ 40</math> | ||
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==Solution 2== | ==Solution 2== | ||
− | + | To make the sum the greatest, put the three largest numbers <math>(13,14</math> and <math>15)</math> in the corners. Then, balance the sides by putting the least integer <math>(10)</math> between the greatest sum <math>(14</math> and <math>15)</math>. Then put the next least integer <math>(11)</math> between the next greatest sum (<math>13 +15</math>). Fill in the last integer <math>(12)</math> and you can see that the sum of any three numbers on a side is (for example) <math>14 +10 + 15 = 39</math> | |
<math>\boxed{\text{D}}</math>. | <math>\boxed{\text{D}}</math>. | ||
-by goldenn | -by goldenn |
Latest revision as of 10:05, 10 June 2024
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 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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.