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
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 .
See Also
[[{{{year}}} AMC 8]] ([[{{{year}}} AMC 8 Problems|Problems]] • [[{{{year}}} AMC 8 Answer Key|Answer Key]] • Resources) | ||
Preceded by [[{{{year}}} AMC 8 Problems/Problem 20|Problem 20]] |
Followed by [[{{{year}}} AMC 8 Problems/Problem 22|Problem 22]] | |
[[{{{year}}} AMC 8 Problems/Problem 1|1]] • [[{{{year}}} AMC 8 Problems/Problem 2|2]] • [[{{{year}}} AMC 8 Problems/Problem 3|3]] • [[{{{year}}} AMC 8 Problems/Problem 4|4]] • [[{{{year}}} AMC 8 Problems/Problem 5|5]] • [[{{{year}}} AMC 8 Problems/Problem 6|6]] • [[{{{year}}} AMC 8 Problems/Problem 7|7]] • [[{{{year}}} AMC 8 Problems/Problem 8|8]] • [[{{{year}}} AMC 8 Problems/Problem 9|9]] • [[{{{year}}} AMC 8 Problems/Problem 10|10]] • [[{{{year}}} AMC 8 Problems/Problem 11|11]] • [[{{{year}}} AMC 8 Problems/Problem 12|12]] • [[{{{year}}} AMC 8 Problems/Problem 13|13]] • [[{{{year}}} AMC 8 Problems/Problem 14|14]] • [[{{{year}}} AMC 8 Problems/Problem 15|15]] • [[{{{year}}} AMC 8 Problems/Problem 16|16]] • [[{{{year}}} AMC 8 Problems/Problem 17|17]] • [[{{{year}}} AMC 8 Problems/Problem 18|18]] • [[{{{year}}} AMC 8 Problems/Problem 19|19]] • [[{{{year}}} AMC 8 Problems/Problem 20|20]] • [[{{{year}}} AMC 8 Problems/Problem 21|21]] • [[{{{year}}} AMC 8 Problems/Problem 22|22]] • [[{{{year}}} AMC 8 Problems/Problem 23|23]] • [[{{{year}}} AMC 8 Problems/Problem 24|24]] • [[{{{year}}} AMC 8 Problems/Problem 25|25]] | ||
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