Difference between revisions of "1999 AMC 8 Problems/Problem 11"

Latest revision as of 17:23, 7 November 2020

Problem

Each of the five numbers 1, 4, 7, 10, and 13 is placed in one of the five squares so that the sum of the three numbers in the horizontal row equals the sum of the three numbers in the vertical column. The largest possible value for the horizontal or vertical sum is

$[asy] draw((0,0)--(3,0)--(3,1)--(0,1)--cycle); draw((1,-1)--(2,-1)--(2,2)--(1,2)--cycle); [/asy]$

$\text{(A)}\ 20 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 22 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 30$

Solution

Solution 1

The largest sum occurs when $13$ is placed in the center. This sum is $13 + 10 + 1 = 13 + 7 + 4 = \boxed{\text{(D)}\ 24}$ . Note: Two other common sums, $18$ and $21$ , are also possible.

Solution 2

Since the horizontal sum equals the vertical sum, twice this sum will be the sum of the five numbers plus the number in the center. When the center number is $13$ , the sum is the largest, $[10 + 4 + 1 + 7 + 2(13)]=2S\\ 48=2S\\ S=\boxed{\text{(D)}\ 24}$ The other four numbers are divided into two pairs with equal sums.

1999 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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.

@@ Line 1: / Line 1: @@
-Problem 11
+==Problem==
 Each of the five numbers 1, 4, 7, 10, and 13 is placed in one of the five squares so that the sum of the three numbers in the horizontal row equals the sum of the three numbers in the vertical column. The largest possible value for the horizontal or vertical sum is
+<asy>
+draw((0,0)--(3,0)--(3,1)--(0,1)--cycle);
+draw((1,-1)--(2,-1)--(2,2)--(1,2)--cycle);
+</asy>
+<math>\text{(A)}\ 20 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 22 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 30</math>
+==Solution==
+===Solution 1===
+The largest sum occurs when <math>13</math> is placed in the center. This sum is <math>13 + 10 + 1 = 13 + 7 + 4 = \boxed{\text{(D)}\ 24}</math>. Note: Two other common sums, <math>18</math> and <math>21</math>, are also possible.
+===Solution 2===
-[asy]
+Since the horizontal sum equals the vertical sum, twice this sum will be the sum
-draw((0,0)--(3,0)--(3,1)--(0,1)--cycle);
+of the five numbers plus the number in the center. When the center number is
-draw((1,-1)--(2,-1)--(2,2)--(1,2)--cycle);[/asy]
+<math>13</math>, the sum is the largest, <cmath>[10 + 4 + 1 + 7 + 2(13)]=2S\\ 48=2S\\ S=\boxed{\text{(D)}\ 24}</cmath>
+The other
+four numbers are divided into two pairs with equal sums.
+==See Also==
-The answer is obviously (D) because you have to put the biggest number (13) in the middle. Then, in order to make it equal, you put the next biggest number (10) on the top, and then you put 7 on the left side. Since the difference for the numbers are all 3, you have to put the next biggest one (4) on the horizontal row, and then you put the (1) on the vertical row, and get letter (D) as your answer.
+{{AMC8 box|year=1999|num-b=10|num-a=12}}
+{{MAA Notice}}

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