Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AMC 10A Problems/Problem 7"

m (Removed a name that was in the wrong place.)
 
(18 intermediate revisions by 14 users not shown)
Line 3: Line 3:
 
==Problem==
 
==Problem==
 
The <math>25</math> integers from <math>-10</math> to <math>14,</math> inclusive, can be arranged to form a <math>5</math>-by-<math>5</math> square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
 
The <math>25</math> integers from <math>-10</math> to <math>14,</math> inclusive, can be arranged to form a <math>5</math>-by-<math>5</math> square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
 +
 
<math>\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50</math>
 
<math>\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50</math>
  
−
== Solution ==  
+
== Solution 1 ==  
  
−
Without loss of generality, consider the five rows in the square. Each row must have the same sum of numbers, meaning that the sum of all the numbers in the square divided by <math>5</math> is the total value per row. The sum of the <math>25</math> integers is <math>-10+9+...+14=11+12+13+14=50</math>, and the common sum is <math>\frac{50}{5}=\boxed{\text{(C) }10}</math>.
+
Without loss of generality, consider the five rows in the square. Each row must have the same sum of numbers, meaning that the sum of all the numbers in the square divided by <math>5</math> is the total value per row. The sum of the <math>25</math> integers is <math>-10+-9+...+14=11+12+13+14=50</math>, and the common sum is <math>\frac{50}{5}=\boxed{\text{(C) }10}</math>.
  
−
 
+
==Solution 2==
−
===Solution 2===
 
  
 
Take the sum of the middle 5 values of the set (they will turn out to be the mean of each row). We get <math>0 + 1 + 2 + 3 + 4 = \boxed{\textbf{(C) } 10}</math> as our answer.
 
Take the sum of the middle 5 values of the set (they will turn out to be the mean of each row). We get <math>0 + 1 + 2 + 3 + 4 = \boxed{\textbf{(C) } 10}</math> as our answer.
 
~Baolan
 
~Baolan
  
−
 
+
==Solution 3==
−
===Solution 3===
 
  
 
Taking the average of the first and last terms, <math>-10</math> and <math>14</math>, we have that the mean of the set is <math>2</math>. There are 5 values in each row, column or diagonal, so the value of the common sum is <math>5\cdot2</math>, or <math>\boxed{\textbf{(C) } 10}</math>.  
 
Taking the average of the first and last terms, <math>-10</math> and <math>14</math>, we have that the mean of the set is <math>2</math>. There are 5 values in each row, column or diagonal, so the value of the common sum is <math>5\cdot2</math>, or <math>\boxed{\textbf{(C) } 10}</math>.  
 
~Arctic_Bunny, edited by KINGLOGIC
 
~Arctic_Bunny, edited by KINGLOGIC
  
−
==Video Solution==
+
==Solution 4==
 +
 
 +
Let us consider the horizontal rows.  Since there are five of them, each with constant sum <math>x</math>, we can add up the 25 numbers in 5 rows for a sum of <math>5x</math>.  Since the sum of the 25 numbers used is <math>-10-9-8-\cdots{}+12+13+14+15=11+12+13+14+15=50</math>, <math>5x=50</math> and <math>x=\boxed{\textbf{(C) }10}</math>.
 +
~cw357
 +
 
 +
==Solution 5==
 +
 
 +
The mean of the set of numbers is <math>(14-10) \div 2 = 2</math>. The numbers around it must be equal (i.e. if the mean of <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</math> is <math>3</math>, then <math>2+4=1+5</math>.)
 +
One row of the square would be
 +
<cmath>\square  \square  2  \square  \square </cmath>
 +
 
 +
Adding the numbers would be
 +
 
 +
<cmath> 0, 1, 2, 3, 4</cmath>
 +
 
 +
with a sum of <math>\boxed {\textbf {(C) }10}</math>.
 +
 
 +
<!-- Very bad and trivial solution :( -->
 +
 
 +
==Solution 6==
 +
If the sum of each row, column, and diagonal is x, then we have a total of 12x for the sum. The sum of the rows and columns is the sum of all the numbers doubled, which is <math>50\cdot2=100</math>. Therefore <math>100+2x=12x</math>, <math>100=10x</math>, and <math>x=\boxed{10}</math>.
 +
~MC413551
 +
 
 +
==Video Solution 1==
 +
 
 +
Education, the Study of Everything
 +
 
 +
https://youtu.be/Zf4HCY-y5Z4
 +
 
 +
==Video Solution 2==
 
https://youtu.be/JEjib74EmiY
 
https://youtu.be/JEjib74EmiY
  
 
~IceMatrix
 
~IceMatrix
 +
 +
==Video Solution 3==
 +
https://youtu.be/PHHBIiIlCY0
 +
 +
~savannahsolver
 +
 +
== Video Solution 4 by OmegaLearn==
 +
https://youtu.be/mgEZOXgIZXs?t=1
 +
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Latest revision as of 13:02, 30 September 2023

The following problem is from both the 2020 AMC 12A #5 and 2020 AMC 10A #7, so both problems redirect to this page.

Problem

The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$-by-$5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?

$\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$

Solution 1

Without loss of generality, consider the five rows in the square. Each row must have the same sum of numbers, meaning that the sum of all the numbers in the square divided by $5$ is the total value per row. The sum of the $25$ integers is $-10+-9+...+14=11+12+13+14=50$, and the common sum is $\frac{50}{5}=\boxed{\text{(C) }10}$.

Solution 2

Take the sum of the middle 5 values of the set (they will turn out to be the mean of each row). We get $0 + 1 + 2 + 3 + 4 = \boxed{\textbf{(C) } 10}$ as our answer. ~Baolan

Solution 3

Taking the average of the first and last terms, $-10$ and $14$, we have that the mean of the set is $2$. There are 5 values in each row, column or diagonal, so the value of the common sum is $5\cdot2$, or $\boxed{\textbf{(C) } 10}$. ~Arctic_Bunny, edited by KINGLOGIC

Solution 4

Let us consider the horizontal rows. Since there are five of them, each with constant sum $x$, we can add up the 25 numbers in 5 rows for a sum of $5x$. Since the sum of the 25 numbers used is $-10-9-8-\cdots{}+12+13+14+15=11+12+13+14+15=50$, $5x=50$ and $x=\boxed{\textbf{(C) }10}$. ~cw357

Solution 5

The mean of the set of numbers is $(14-10) \div 2 = 2$. The numbers around it must be equal (i.e. if the mean of $1$, $2$, $3$, $4$, and $5$ is $3$, then $2+4=1+5$.) One row of the square would be \[\square \square 2 \square \square\]

Adding the numbers would be

\[0, 1, 2, 3, 4\]

with a sum of $\boxed {\textbf {(C) }10}$.


Solution 6

If the sum of each row, column, and diagonal is x, then we have a total of 12x for the sum. The sum of the rows and columns is the sum of all the numbers doubled, which is $50\cdot2=100$. Therefore $100+2x=12x$, $100=10x$, and $x=\boxed{10}$. ~MC413551

Video Solution 1

Education, the Study of Everything

https://youtu.be/Zf4HCY-y5Z4

Video Solution 2

https://youtu.be/JEjib74EmiY

~IceMatrix

Video Solution 3

https://youtu.be/PHHBIiIlCY0

~savannahsolver

Video Solution 4 by OmegaLearn

https://youtu.be/mgEZOXgIZXs?t=1

~ pi_is_3.14

See Also

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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 AMC 10 Problems and Solutions
2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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 AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_7&oldid=198881"