2014 AMC 8 Problems/Problem 8: Difference between revisions
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==Solution== | ==Solution== | ||
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>. | 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>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2014|num-b=7|num-a=9}} | {{AMC8 box|year=2014|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:42, 3 November 2015
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 
. What is the missing digit 
of this 
-digit number?
Solution
A number is divisible by 
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 . So 
is equivalent to 
. Clearly 3 , (A) cannot be equal to or any multiple of greater than that. So 
, or 
.
See Also
| 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
