Difference between revisions of "2022 AMC 8 Problems/Problem 20"

Latest revision as of 22:35, 29 December 2023

1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4 (Answer Choices)
6 Solution 5 (Super fast! No algebra; no testing answer choices)
7 Video Solution by Math-X (First understand the problem!!!)
8 Video Solution (🚀Super Fast. Just 1 min!🚀)
9 Video Solution
10 Video Solution
11 Video Solution
12 Video Solution
13 Video Solution
14 Video Solution
15 See Also

Problem

The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ ? $[asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$x$"); [/asy]$ $\textbf{(A) } -1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9 \qquad$

Solution 1

The sum of the numbers in each row is $12$ . Consider the second row. In order for the sum of the numbers in this row to equal $12$ , the two shaded numbers must add up to $13$ : $[asy] unitsize(0.5cm); fill((-3,1)--(1,1)--(1,-1)--(-3,-1)--cycle,mediumgray); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$x$"); [/asy]$ If two numbers add up to $13$ , one of them must be at least $7$ : If both shaded numbers are no more than $6$ , their sum can be at most $12$ . Therefore, for $x$ to be larger than the three missing numbers, $x$ must be at least $8$ . We can construct a working scenario where $x=8$ : $[asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$8$"); label((0,-2),"$-4$"); label((-2,0),"$6$"); label((0,0),"$7$"); [/asy]$ So, our answer is $\boxed{\textbf{(D) } 8}$ .

~ihatemath123

Solution 2

The sum of the numbers in each row is $-2+9+5=12,$ and the sum of the numbers in each column is $5+(-1)+8=12.$

Let $y$ be the number in the lower middle. It follows that $x+y+8=12,$ or $x+y=4.$

We express the other two missing numbers in terms of $x$ and $y,$ as shown below: $[asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$x$"); label((0,-2),"$y$",red+fontsize(11)); label((-2,0),"$y{+}10$",red+fontsize(11)); label((0,0),"$x{-}1$",red+fontsize(11)); [/asy]$ We have $x>x-1, x>y+10,$ and $x>y.$ Note that the first inequality is true for all values of $x.$ We only need to solve the second inequality so that the third inequality is true for all values of $x.$ By substitution, we get $x>(4-x)+10,$ from which $x>7.$

Therefore, the smallest possible value of $x$ is $\boxed{\textbf{(D) } 8}.$

~MRENTHUSIASM

Solution 3

This is based on the Solution 2 above and it is perhaps a little simpler than that.

Let $y$ be the number in the lower middle. Applying summation to first two columns yields the following.

$[asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$x$"); label((0,-2),"$y$",red+fontsize(11)); label((-2,0),"$14{-}x$",red+fontsize(11)); label((0,0),"$3{-}y$",red+fontsize(11)); [/asy]$

Since $x$ is greater than the other three, we have $x>14-x,$ or $x>7.$

Therefore, the smallest possible value of $x$ is $\boxed{\textbf{(D) } 8}.$

~vetaltekdi6

Solution 4 (Answer Choices)

Note that the sum of the rows and columns must be $8+5-1=12$ . We proceed to test the answer choices.

Testing $\textbf{(A)}$ , when $x = -1$ , the number above $x$ must be $15$ , which contradicts the precondition that the numbers surrounding $x$ is less than $x$ .

Testing $\textbf{(B)}$ , the number above $x$ is $9$ , which does not work.

Testing $\textbf{(C)}$ , the number above $x$ is $8$ , which does not work.

Testing $\textbf{(D)}$ , the number above $x$ is $6$ , which does work. Hence, the answer is $\boxed{\textbf{(D) }8}$ .

We do not need to test $\textbf{(E)}$ , because the problem asks for the smallest value of $x$ .

~MrThinker

Solution 5 (Super fast! No algebra; no testing answer choices)

The sum of the numbers in each column and row should be $5+(-1)+8=12$ . If we look at the $1^{\text{st}}$ column, the gray squares (shown below) sum to $12-(-2)=14$ .

$[asy] draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$x$"); filldraw((-3,-3)--(-1,-3)--(-1,-1)--(-3,-1)--cycle, lightgray, black+linewidth(1)); filldraw((-1,-1)--(-3,-1)--(-3,1)--(-1,1)--cycle, lightgray, black+linewidth(1)); label(scale(1)*"All credits for original unedited asymptote for the problem go to whoever made the asymptote in the 'Problem' section.", (-0,-5), S); [/asy]$

If square $x$ has to be greater than or equal to the three blank squares, then the least $x$ can be is half the sum of the value of the gray squares, which is $14\div2=7$ . But square $x$ has to be greater than and not greater than or equal to the three blank squares, so the least $x$ can be is $7+1=8$ . Testing for the other rows and columns (it might be smaller than the other two squares!), we find that the smallest $x$ can be is indeed $8$ ; the other two squares are less than $8$ . Therefore, the answer is $\boxed{\text{(D) }8}$

~ JoyfulSapling

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/oUEa7AjMF2A?si=Bbea8RWE2sMWN6xl&t=3643

~Math-X

Video Solution (🚀Super Fast. Just 1 min!🚀)

https://youtu.be/7J4EGPaB29Y

~Education, the Study of Everything

Video Solution

https://youtu.be/0hHlpIVeFjg

Please like and subscribe!

Video Solution

https://www.youtube.com/watch?v=xnGQffaxYAA

~Mathematical Dexterity

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=1857

~Interstigation

Video Solution

https://youtu.be/hs6y4PWnoWg?t=369

~STEMbreezy

Video Solution

https://youtu.be/DXFwzrOF4b4

~savannahsolver

Video Solution

https://www.youtube.com/watch?v=FINk9LgSJpU

~David

2022 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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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Difference between revisions of "2022 AMC 8 Problems/Problem 20"

Latest revision as of 22:35, 29 December 2023

Contents

Problem

Solution 1

Solution 2

Solution 3

Solution 4 (Answer Choices)

Solution 5 (Super fast! No algebra; no testing answer choices)

Video Solution by Math-X (First understand the problem!!!)

Video Solution (🚀Super Fast. Just 1 min!🚀)

Video Solution

Video Solution

Video Solution

Video Solution

Video Solution

Video Solution

See Also

@@ Line 1: / Line 1: @@
-==Problem==
+== Problem ==
 The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number <math>x</math> in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of <math>x</math>?
@@ Line 15: / Line 15: @@
 label((0,2),"$9$");
 label((2,2),"$5$");
-label((2,0),"${-}1$");
+label((2,0),"$-1$");
 label((2,-2),"$8$");
 label((-2,-2),"$x$");
 </asy>
-<math>\textbf{(A) } {-}1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9</math>
+<math>\textbf{(A) } -1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9 \qquad</math>
-==Solution==
+== Solution 1 ==
+The sum of the numbers in each row is <math>12</math>. Consider the second row. In order for the sum of the numbers in this row to equal <math>12</math>, the two shaded numbers must add up to <math>13</math>:
+<asy>
+unitsize(0.5cm);
+fill((-3,1)--(1,1)--(1,-1)--(-3,-1)--cycle,mediumgray);
+draw((3,3)--(-3,3));
+draw((3,1)--(-3,1));
+draw((3,-3)--(-3,-3));
+draw((3,-1)--(-3,-1));
+draw((3,3)--(3,-3));
+draw((1,3)--(1,-3));
+draw((-3,3)--(-3,-3));
+draw((-1,3)--(-1,-3));
+label((-2,2),"$-2$");
+label((0,2),"$9$");
+label((2,2),"$5$");
+label((2,0),"$-1$");
+label((2,-2),"$8$");
+label((-2,-2),"$x$");
+</asy>
+If two numbers add up to <math>13</math>, one of them must be at least <math>7</math>: If both shaded numbers are no more than <math>6</math>, their sum can be at most <math>12</math>. Therefore, for <math>x</math> to be larger than the three missing numbers, <math>x</math> must be at least <math>8</math>. We can construct a working scenario where <math>x=8</math>:
+<asy>
+unitsize(0.5cm);
+draw((3,3)--(-3,3));
+draw((3,1)--(-3,1));
+draw((3,-3)--(-3,-3));
+draw((3,-1)--(-3,-1));
+draw((3,3)--(3,-3));
+draw((1,3)--(1,-3));
+draw((-3,3)--(-3,-3));
+draw((-1,3)--(-1,-3));
+label((-2,2),"$-2$");
+label((0,2),"$9$");
+label((2,2),"$5$");
+label((2,0),"$-1$");
+label((2,-2),"$8$");
+label((-2,-2),"$8$");
+label((0,-2),"$-4$");
+label((-2,0),"$6$");
+label((0,0),"$7$");
+</asy>
+So, our answer is <math>\boxed{\textbf{(D) } 8}</math>.
+~ihatemath123
+==Solution 2==
+The sum of the numbers in each row is <math>-2+9+5=12,</math> and the sum of the numbers in each column is <math>5+(-1)+8=12.</math>
+Let <math>y</math> be the number in the lower middle. It follows that <math>x+y+8=12,</math> or <math>x+y=4.</math>
+We express the other two missing numbers in terms of <math>x</math> and <math>y,</math> as shown below:
+<asy>
+unitsize(0.5cm);
+draw((3,3)--(-3,3));
+draw((3,1)--(-3,1));
+draw((3,-3)--(-3,-3));
+draw((3,-1)--(-3,-1));
+draw((3,3)--(3,-3));
+draw((1,3)--(1,-3));
+draw((-3,3)--(-3,-3));
+draw((-1,3)--(-1,-3));
+label((-2,2),"$-2$");
+label((0,2),"$9$");
+label((2,2),"$5$");
+label((2,0),"$-1$");
+label((2,-2),"$8$");
+label((-2,-2),"$x$");
+label((0,-2),"$y$",red+fontsize(11)); label((-2,0),"$y{+}10$",red+fontsize(11)); label((0,0),"$x{-}1$",red+fontsize(11));
+</asy>
+We have <math>x>x-1, x>y+10,</math> and <math>x>y.</math> Note that the first inequality is true for all values of <math>x.</math> We only need to solve the second inequality so that the third inequality is true for all values of <math>x.</math> By substitution, we get <math>x>(4-x)+10,</math> from which <math>x>7.</math>
+Therefore, the smallest possible value of <math>x</math> is <math>\boxed{\textbf{(D) } 8}.</math>
+~MRENTHUSIASM
+==Solution 3==
+This is based on the Solution 2 above and it is perhaps a little simpler than that.
+Let <math>y</math> be the number in the lower middle. Applying summation to first two columns yields the following.
+<asy>
+unitsize(0.5cm);
+draw((3,3)--(-3,3));
+draw((3,1)--(-3,1));
+draw((3,-3)--(-3,-3));
+draw((3,-1)--(-3,-1));
+draw((3,3)--(3,-3));
+draw((1,3)--(1,-3));
+draw((-3,3)--(-3,-3));
+draw((-1,3)--(-1,-3));
+label((-2,2),"$-2$");
+label((0,2),"$9$");
+label((2,2),"$5$");
+label((2,0),"$-1$");
+label((2,-2),"$8$");
+label((-2,-2),"$x$");
+label((0,-2),"$y$",red+fontsize(11)); label((-2,0),"$14{-}x$",red+fontsize(11)); label((0,0),"$3{-}y$",red+fontsize(11));
+</asy>
+Since <math>x</math> is greater than the other three, we have <math>x>14-x,</math> or <math>x>7.</math>
+Therefore, the smallest possible value of <math>x</math> is <math>\boxed{\textbf{(D) } 8}.</math>
+~vetaltekdi6
+==Solution 4 (Answer Choices)==
+Note that the sum of the rows and columns must be <math>8+5-1=12</math>. We proceed to test the answer choices.
+Testing <math>\textbf{(A)}</math>, when <math>x = -1</math>, the number above <math>x</math> must be <math>15</math>, which contradicts the precondition that the numbers surrounding <math>x</math> is less than <math>x</math>.
+Testing <math>\textbf{(B)}</math>, the number above <math>x</math> is <math>9</math>, which does not work.
+Testing <math>\textbf{(C)}</math>, the number above <math>x</math> is <math>8</math>, which does not work.
+Testing <math>\textbf{(D)}</math>, the number above <math>x</math> is <math>6</math>, which ''does'' work. Hence, the answer is <math>\boxed{\textbf{(D) }8}</math>.
+We do not need to test <math>\textbf{(E)}</math>, because the problem asks for the '''smallest''' value of <math>x</math>.
+~MrThinker
+==Solution 5 (Super fast! No algebra; no testing answer choices)==
+The sum of the numbers in each column and row should be <math>5+(-1)+8=12</math>. If we look at the <math>1^{\text{st}}</math> column, the gray squares (shown below) sum to <math>12-(-2)=14</math>.
+<asy>
+draw((3,3)--(-3,3));
+draw((3,1)--(-3,1));
+draw((3,-3)--(-3,-3));
+draw((3,-1)--(-3,-1));
+draw((3,3)--(3,-3));
+draw((1,3)--(1,-3));
+draw((-3,3)--(-3,-3));
+draw((-1,3)--(-1,-3));
+label((-2,2),"$-2$");
+label((0,2),"$9$");
+label((2,2),"$5$");
+label((2,0),"$-1$");
+label((2,-2),"$8$");
+label((-2,-2),"$x$");
+filldraw((-3,-3)--(-1,-3)--(-1,-1)--(-3,-1)--cycle, lightgray, black+linewidth(1));
+filldraw((-1,-1)--(-3,-1)--(-3,1)--(-1,1)--cycle, lightgray, black+linewidth(1));
+label(scale(1)*"All credits for original unedited asymptote for the problem go to whoever made the asymptote in the 'Problem' section.", (-0,-5), S);
+</asy>
+'''If''' square <math>x</math> has to be '''greater than or equal to''' the three blank squares, then the least <math>x</math> can be is half the sum of the value of the gray squares, which is <math>14\div2=7</math>. But square <math>x</math> has to be '''greater than''' and '''''not''''' '''greater than or equal to''' the three blank squares, so the least <math>x</math> can be is <math>7+1=8</math>. Testing for the other rows and columns (it might be smaller than the other two squares!), we find that the smallest <math>x</math> can be is indeed <math>8</math>; the other two squares are less than <math>8</math>. Therefore, the answer is <math>\boxed{\text{(D) }8}</math>
+~ JoyfulSapling
+==Video Solution by Math-X (First understand the problem!!!)==
+https://youtu.be/oUEa7AjMF2A?si=Bbea8RWE2sMWN6xl&t=3643
+~Math-X
+==Video Solution (🚀Super Fast. Just 1 min!🚀)==
+https://youtu.be/7J4EGPaB29Y
+<i>~Education, the Study of Everything</i>
+==Video Solution==
+https://youtu.be/0hHlpIVeFjg
+Please like and subscribe!
+==Video Solution==
+https://www.youtube.com/watch?v=xnGQffaxYAA
+~Mathematical Dexterity
+==Video Solution==
+https://youtu.be/Ij9pAy6tQSg?t=1857
+~Interstigation
+==Video Solution==
+https://youtu.be/hs6y4PWnoWg?t=369
+~STEMbreezy
+==Video Solution==
+https://youtu.be/DXFwzrOF4b4
+~savannahsolver
+==Video Solution==
+https://www.youtube.com/watch?v=FINk9LgSJpU
+~David
 ==See Also==
 {{AMC8 box|year=2022|num-b=19|num-a=21}}
 {{MAA Notice}}