Difference between revisions of "1985 AJHSME Problems/Problem 24"
(→See Also) |
|||
(14 intermediate revisions by 9 users not shown) | |||
Line 20: | Line 20: | ||
<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> | ||
− | ==Solution== | + | ==Solution 1== |
− | + | 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>. | |
− | + | -by goldenn | |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
+ | ==Solution 2== | ||
+ | Let the numbers, in clockwise order from the very top, be <math>a, b, c, d, e, f</math>. Notice that <math>a + b + c + d + e + f = \frac{15(16)}{2} - \frac{9(10)}{2}</math>. Manipulation will deal <math>3S - c - e - f = 75</math>. Now we use the answer choices to our advantage. We first check for <math>S = 40</math> and find that <math>c + e + f = 45</math>, which is impossible due to the restrictions in the question. Having eliminated 40, we now move onto <math>S = 39</math> and find that this deals <math>c + e + f = 42</math>, which is possible (for example, let <math>c = 13, e = 14, f = 15</math>). Thus, the answer is <math>\textbf{(D)} 39</math>. | ||
+ | -scthecool | ||
==See Also== | ==See Also== | ||
Latest revision as of 14:19, 16 October 2024
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 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 | |
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.