Difference between revisions of "2024 AMC 8 Problems/Problem 4"
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<math>\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9</math> | <math>\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9</math> | ||
==Solution 1== | ==Solution 1== | ||
− | The sum of the numbers from <math>1</math> to <math>9</math> is <cmath>1 + 2 + 3 + \cdots + 9 = \frac{9(10)}{2} = 45.</cmath> Denote the number left out when adding to be <math>x</math>. Thus, <math>45 - x</math> is a perfect square. We could rigorously solve the values of <math>x</math> such that <math>45 - x</math> is a perfect square. However, since this is the AMC 8, we can simply try each option. When <math>x = 9</math>, we see that <math>45 - 9 = 36</math>, a perfect square. Therefore, our answer is <math>\boxed{\textbf{(E) 9 | + | The sum of the numbers from <math>1</math> to <math>9</math> is <cmath>1 + 2 + 3 + \cdots + 9 = \frac{9(10)}{2} = 45.</cmath> Denote the number left out when adding to be <math>x</math>. Thus, <math>45 - x</math> is a perfect square. We could rigorously solve the values of <math>x</math> such that <math>45 - x</math> is a perfect square. However, since this is the AMC 8, we can simply try each option. When <math>x = 9</math>, we see that <math>45 - 9 = 36</math>, a perfect square. Therefore, our answer is <math>\boxed{\textbf{(E) }9}</math>. |
-rnatog337 | -rnatog337 |
Revision as of 14:03, 25 January 2024
Problem
When Yunji added all the integers from to
, she mistakenly left out a number. Her incorrect sum turned out to be a square number. What number did Yunji leave out?
Solution 1
The sum of the numbers from to
is
Denote the number left out when adding to be
. Thus,
is a perfect square. We could rigorously solve the values of
such that
is a perfect square. However, since this is the AMC 8, we can simply try each option. When
, we see that
, a perfect square. Therefore, our answer is
.
-rnatog337