2014 AMC 8 Problems/Problem 21
Problem
The 
-digit numbers 
and 
are each multiples of 
. Which of the following could be the value of 
?
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. 7 + 4 + 5 + 2 + 1 = 19. In order to be a multiple of 3, A + B has to be either 5 or 8. We add up the numerical digits in the second number; 3 + 2 + 6 + 4 = 15. We then add 5 to 15, to get 20. We then see that C = 1 or 7, otherwise the number will not be divisible by three. We then add 8 to 15, to get 23, which shows us that C = 1 or 4 or 7 in order to be a multiple of three. We take the common numbers we got from both these equations, which are 1 and 7. However, in the answer choices, there is no 7, but there is a 1, so is your answer. :)
~ UnicornFrappacino
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
Another solution is using the divisbility rule of 3. Just add the digits of each number together.
