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 1
The number is congruent to 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.
Video Solution for Problems 21-25
https://www.youtube.com/watch?v=6S0u_fDjSxc
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 |
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