2014 AMC 8 Problems/Problem 21
Contents
[hide]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 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 | |
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