2014 AMC 8 Problems/Problem 21
Problem
The 
-digit numbers 
and 
are each multiples of 
. Which of the following could be the value of 
?
Video Solution
https://youtu.be/6xNkyDgIhEE?t=2593
Solution 1
The sum of a number's digits
is congruent to the number. 
must be congruent to 0, since it is divisible by 3. Therefore, 
is also congruent to 0. 
, so 
. As we know, 
, so 
, and therefore 
. We can substitute 2 for 
, so 
, and therefore 
. This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is 
.
Solution 2
Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. 
. To be a multiple of , 
has to be either 
or 
or 
... and so on. We add up the numerical digits in the second number; 
. We then add two of the selected values, to 
, to get 
. We then see that C = 
or 
... and so on, otherwise the number will not be divisible by three. We then add to , to get 
, which shows us that C = 
or 
or ... and so on. To be a multiple of three, we select a few of the common numbers we got from both these equations, which could be 
and . However, in the answer choices, there is no or or anything greater than , but there is a , so is our answer.
See Also
| 2014 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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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