Difference between revisions of "2014 AMC 8 Problems/Problem 21"
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The sum of a number's digits <math>\mod{3}</math> is congruent to the number <math>\pmod{3}</math>. <math>74A52B1 \mod{3}</math> must be congruent to 0, since it is divisible by 3. Therefore, <math>7+4+A+5+2+B+1 \mod{3}</math> is also congruent to 0. <math>7+4+5+2+1 \equiv 1 \pmod{3}</math>, so <math>A+B\equiv 2 \pmod{3}</math>. As we know, <math>326AB4C\equiv 0 \pmod{3}</math>, so <math>3+2+6+A+B+4+C =15+A+B+C\equiv 0 \pmod{3}</math>, and therefore <math>A+B+C\equiv 0 \pmod{3}</math>. We can substitute 2 for <math>A+B</math>, so <math>2+C\equiv 0 \pmod{3}</math>, and therefore <math>C\equiv 1\pmod{3}</math>. This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is <math>\textbf{(A) }1</math>. | The sum of a number's digits <math>\mod{3}</math> is congruent to the number <math>\pmod{3}</math>. <math>74A52B1 \mod{3}</math> must be congruent to 0, since it is divisible by 3. Therefore, <math>7+4+A+5+2+B+1 \mod{3}</math> is also congruent to 0. <math>7+4+5+2+1 \equiv 1 \pmod{3}</math>, so <math>A+B\equiv 2 \pmod{3}</math>. As we know, <math>326AB4C\equiv 0 \pmod{3}</math>, so <math>3+2+6+A+B+4+C =15+A+B+C\equiv 0 \pmod{3}</math>, and therefore <math>A+B+C\equiv 0 \pmod{3}</math>. We can substitute 2 for <math>A+B</math>, so <math>2+C\equiv 0 \pmod{3}</math>, and therefore <math>C\equiv 1\pmod{3}</math>. This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is <math>\textbf{(A) }1</math>. | ||
==See Also== | ==See Also== | ||
| − | {{AMC8 box| | + | {{AMC8 box|month=2014|num-b=20|num-a=22}} |
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:47, 5 September 2015
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]] | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
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