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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==
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==Solution 1==
  
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Let the number in the top circle be <math>a</math> and then <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math>, and <math>f</math>, going in clockwise order. Then, we have <cmath>S=a+b+c</cmath> <cmath>S=c+d+e</cmath> <cmath>S=e+f+a</cmath>
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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>
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Adding these [[equation|equations]] together, we get
 
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<cmath>\begin{align*}
 
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3S &= (a+b+c+d+e+f)+(a+c+e) \\
 
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&= 75+(a+c+e) \\
 
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\end{align*}</cmath>
 
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where the last step comes from the fact that since <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math>, and <math>f</math> are the numbers <math>10-15</math> in some order, their [[sum]] is <math>10+11+12+13+14+15=75</math>
 
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The left hand side is [[divisible]] by <math>3</math> and <math>75</math> is divisible by <math>3</math>, so <math>a+c+e</math> must be divisible by <math>3</math>.  The largest possible value of <math>a+c+e</math> is then <math>15+14+13=42</math>, and the corresponding value of <math>S</math> is <math>\frac{75+42}{3}=39</math>, which is choice <math>\boxed{\text{D}}</math>.
 
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It turns out this sum is attainable if you let <cmath>a=15</cmath> <cmath>b=10</cmath> <cmath>c=14</cmath> <cmath>d=12</cmath> <cmath>e=13</cmath> <cmath>f=11</cmath>
 
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==Solution 2==
 
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TO make the sum the greatest, put the three largest numbers (13,14,and 15) in the corners. Then, balance the sides by putting the least integer (<math>10</math>) between the greatest sum (14 and 15). Then put the next least integer (<math>11</math>) between the next greatest sum (<math>13 +15</math>). Fill in the last integer(12) and you can see that the sum of any three numbers on a side is (for example) 14 +10 + 15 = <math>39</math>
 
 
<math>\boxed{\text{D}}</math>.  
 
<math>\boxed{\text{D}}</math>.  
 
-by goldenn
 
-by goldenn
  
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==Solution 2==
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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>.
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-scthecool
 
==See Also==
 
==See Also==
  

Latest revision as of 14:19, 16 October 2024

Problem

In a magic triangle, each of the six whole numbers $10-15$ is placed in one of the circles so that the sum, $S$, of the three numbers on each side of the triangle is the same. The largest possible value for $S$ is

[asy] draw(circle((0,0),1)); draw(dir(60)--6*dir(60)); draw(circle(7*dir(60),1)); draw(8*dir(60)--13*dir(60)); draw(circle(14*dir(60),1)); draw((1,0)--(6,0)); draw(circle((7,0),1)); draw((8,0)--(13,0)); draw(circle((14,0),1)); draw(circle((10.5,6.0621778264910705273460621952706),1)); draw((13.5,0.86602540378443864676372317075294)--(11,5.1961524227066318805823390245176)); draw((10,6.9282032302755091741097853660235)--(7.5,11.258330249197702407928401219788)); [/asy]

$\text{(A)}\ 36 \qquad \text{(B)}\ 37 \qquad \text{(C)}\ 38 \qquad \text{(D)}\ 39 \qquad \text{(E)}\ 40$

Solution 1

To make the sum the greatest, put the three largest numbers $(13,14$ and $15)$ in the corners. Then, balance the sides by putting the least integer $(10)$ between the greatest sum $(14$ and $15)$. Then put the next least integer $(11)$ between the next greatest sum ($13 +15$). Fill in the last integer $(12)$ and you can see that the sum of any three numbers on a side is (for example) $14 +10 + 15 = 39$ $\boxed{\text{D}}$. -by goldenn

Solution 2

Let the numbers, in clockwise order from the very top, be $a, b, c, d, e, f$. Notice that $a + b + c + d + e + f = \frac{15(16)}{2} - \frac{9(10)}{2}$. Manipulation will deal $3S - c - e - f = 75$. Now we use the answer choices to our advantage. We first check for $S = 40$ and find that $c + e + f = 45$, which is impossible due to the restrictions in the question. Having eliminated 40, we now move onto $S = 39$ and find that this deals $c + e + f = 42$, which is possible (for example, let $c = 13, e = 14, f = 15$). Thus, the answer is $\textbf{(D)} 39$. -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


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