Difference between revisions of "2014 AMC 8 Problems/Problem 21"
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<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8</math> | <math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8</math> | ||
| − | ==Solution== | + | ==Solution 1== |
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>\boxed{\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>\boxed{\textbf{(A) }1}</math>. | ||
Revision as of 22:09, 12 October 2019
Contents
Problem
The 
-digit numbers 
and 
are each multiples of 
. Which of the following could be the value of 
?
Solution 1
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 
.
Solution 2
Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. 7 + 4 + 5 + 2 + 1 = 19. In order to be a multiple of 3, A + B has to be either 5 or 8. We add up the numerical digits in the second number; 3 + 2 + 6 + 4 = 15. We then add 5 to 15, to get 20. We then see that C = 1 or 7, otherwise the number will not be divisible by three. We then add 8 to 15, to get 23, which shows us that C = 1 or 4 or 7 in order to be a multiple of three. We take the common numbers we got from both these equations, which are 1 and 7. However, in the answer choices, there is no 7, but there is a 1, so is your answer. :)
~ UnicornFrappacino
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. 
Another solution is using the divisbility rule of 3. Just add the digits of each number together.
