Difference between revisions of "1985 AJHSME Problems/Problem 24"

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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>
+
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 17:38, 11 April 2020

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

Let the number in the top circle be $a$ and then $b$, $c$, $d$, $e$, and $f$, going in clockwise order. Then, we have \[S=a+b+c\] \[S=c+d+e\] \[S=e+f+a\]

Adding these equations together, we get

\begin{align*} 3S &= (a+b+c+d+e+f)+(a+c+e) \\ &= 75+(a+c+e) \\ \end{align*}

where the last step comes from the fact that since $a$, $b$, $c$, $d$, $e$, and $f$ are the numbers $10-15$ in some order, their sum is $10+11+12+13+14+15=75$

The left hand side is divisible by $3$ and $75$ is divisible by $3$, so $a+c+e$ must be divisible by $3$. The largest possible value of $a+c+e$ is then $15+14+13=42$, and the corresponding value of $S$ is $\frac{75+42}{3}=39$, which is choice $\boxed{\text{D}}$.

It turns out this sum is attainable if you let \[a=15\] \[b=10\] \[c=14\] \[d=12\] \[e=13\] \[f=11\]


Solution 2

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

See Also

1985 AJHSME (ProblemsAnswer KeyResources)
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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