2014 AMC 8 Problems/Problem 21
Contents
Problem 21
The 
-digit numbers 
and 
are each multiples of 
. Which of the following could be the value of 
?
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.
Concept Solution
for a number to be divisible by 3, the sum of its digits must be divisible by 3. In the number 74A52B1, 7+4+A+5+2+B+1 must be divisible by 3. We get 19+A+B. 21 is the closest multiple of 3 meaning that A=2, B=0, order doesn't matter. Now we plug in those values for 326AB4C. It will be 3+2+6+2+0+4+C to get 17+C. The closest multiple of 3 is 18. So that means 17+C=18. Solving for C, our answer is -TheNerdWhoIsNerdy.
Video Solution by Pi Academy (Fast and Easy ⚡️🚀)
https://youtu.be/t-SjP-gqw20?si=y6ECC_zEYV5OcxY2
Video Solution by OmegaLearn
https://youtu.be/6xNkyDgIhEE?t=2593
Video Solution
https://youtu.be/7TOtBiod55Q ~savannahsolver
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
