Difference between revisions of "1985 AJHSME Problems/Problem 24"
(→Solution 2) |
(→Solution 2) |
||
Line 41: | Line 41: | ||
==Solution 2== | ==Solution 2== | ||
− | TO make the sum the greatest, put the three largest numbers <math>(</math>13,14<math>,and </math>15<math>)</math> in the corners. Then, balance the sides by putting the least integer <math>(10)</math> between the greatest sum <math>(14 and 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> | + | TO make the sum the greatest, put the three largest numbers <math>(</math>13,14<math>,and </math> 15<math>)</math> in the corners. Then, balance the sides by putting the least integer <math>(10)</math> between the greatest sum <math>(14 and 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 |
Revision as of 14:57, 27 December 2019
Contents
[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
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 13,14 15 in the corners. Then, balance the sides by putting the least integer between the greatest sum . 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.