Difference between revisions of "1985 AJHSME Problems/Problem 24"
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+ | ==Problem== | ||
+ | In a magic triangle, each of the six [[whole number|whole numbers]] <math>10-15</math> is placed in one of the [[circle|circles]] so that the sum, <math>S</math>, of the three numbers on each side of the [[triangle]] is the same. The largest possible value for <math>S</math> 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> | ||
+ | |||
+ | <math>\text{(A)}\ 36 \qquad \text{(B)}\ 37 \qquad \text{(C)}\ 38 \qquad \text{(D)}\ 39 \qquad \text{(E)}\ 40</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | 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> | ||
+ | |||
+ | Adding these [[equation|equations]] together, we get | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | 3S &= (a+b+c+d+e+f)+(a+c+e) \ | ||
+ | &= 75+(a+c+e) \ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | |||
+ | |||
+ | ==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>. | ||
+ | -by goldenn | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AJHSME box|year=1985|num-b=23|num-a=25}} | ||
+ | [[Category:Introductory Number Theory Problems]] | ||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | |||
+ | |||
+ | {{MAA Notice}} |
Revision as of 11:28, 9 September 2021
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
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 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.