2014 AMC 8 Problems/Problem 8: Difference between revisions

Latest revision as of 23:21, 17 January 2025

Problem

Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at their math club meeting. They paid their guest speaker $\textdollar\underline{1} \underline{A} \underline{2}$ . What is the missing digit $A$ of this $3$ -digit number?

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad \textbf{(E) }4$

Solution 1

Since all the eleven members paid the same amount, that means that the total must be divisible by $11$ . We can do some trial-and-error to get $A=3$ , so our answer is $\boxed{\textbf{(D)}~3}$ ~SparklyFlowers

Solution 2

We know that a number is divisible by $11$ if the odd digits added together minus the even digits added together (or vice versa) is a multiple of $11$ . Thus, we have $1+2-A$ = a multiple of $11$ . The only multiple that works here is $0$ , as $11 \cdot 0 = 0$ . Thus, $A = \boxed{\textbf{(D)}~3}$ ~fn106068

Solution 3

If you know your multiplication tables up to $12*12$ , you know that the out of the choices provided, the only number with $1A2$ that is divisible by $11$ is $132$ or $12*11$ . So the answer is $\boxed{\textbf{(D)}~3}$ . ~Jupiter10

Video Solution (CREATIVE THINKING)

https://youtu.be/kNHe_IMcMiU

~Education, the Study of Everything

Video Solution

https://youtu.be/mHWTWk-xt0o ~savannahsolver

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

@@ Line 4: / Line 4: @@
 <math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad \textbf{(E) }4</math>
-==Solution==
+==Solution 1==
-A number is divisible by <math>11</math> if the difference between the sum of the digits in the odd-numbered slots (e.g. the ones slot, the hundreds slot, etc.) and the sum in the even-numbered slots (e.g. the tens slot, the thousands slot) is a multiple of <math>11</math>. So <math>1 + 2 - A</math> is equivalent to <math>0\pmod{11}</math>. Clearly 3 , (A) cannot be equal to <math>11</math> or any multiple of <math>11</math> greater than that. So <math> 3 - A = 0 \longrightarrow \boxed{A = 3}</math>, or <math>\boxed{D}</math>.
+Since all the eleven members paid the same amount, that means that the total must be divisible by <math>11</math>. We can do some trial-and-error to get <math>A=3</math>, so our answer is <math>\boxed{\textbf{(D)}~3}</math>
+~SparklyFlowers
+==Solution 2==
+We know that a number is divisible by <math>11</math> if the odd digits added together minus the even digits added together (or vice versa) is a multiple of <math>11</math>. Thus, we have <math>1+2-A</math> = a multiple of <math>11</math>. The only multiple that works here is <math>0</math>, as <math>11 \cdot 0 = 0</math>. Thus, <math>A = \boxed{\textbf{(D)}~3}</math>
+~fn106068
+==Solution 3==
+If you know your multiplication tables up to <math>12*12</math>, you know that the out of the choices provided, the only number with <math>1A2</math> that is divisible by <math>11</math> is <math>132</math> or <math>12*11</math>. So the answer is <math>\boxed{\textbf{(D)}~3}</math>. ~Jupiter10
+==Video Solution (CREATIVE THINKING)==
+https://youtu.be/kNHe_IMcMiU
+~Education, the Study of Everything
+==Video Solution==
+https://youtu.be/mHWTWk-xt0o ~savannahsolver
 ==See Also==
 {{AMC8 box|year=2014|num-b=7|num-a=9}}
 {{MAA Notice}}

Page

Toolbox

Search