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Difference between revisions of "2014 AMC 8 Problems/Problem 21"

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==Solution==
 
==Solution==
 
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>.
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==Solution==
  
 
==See Also==
 
==See Also==

Revision as of 19:29, 29 July 2019

Problem

The $7$-digit numbers $\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}$ and $\underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C}$ are each multiples of $3$. Which of the following could be the value of $C$?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8$

Solution

The sum of a number's digits $\mod{3}$ is congruent to the number $\pmod{3}$. $74A52B1 \mod{3}$ must be congruent to 0, since it is divisible by 3. Therefore, $7+4+A+5+2+B+1 \mod{3}$ is also congruent to 0. $7+4+5+2+1 \equiv 1 \pmod{3}$, so $A+B\equiv 2 \pmod{3}$. As we know, $326AB4C\equiv 0 \pmod{3}$, so $3+2+6+A+B+4+C =15+A+B+C\equiv 0 \pmod{3}$, and therefore $A+B+C\equiv 0 \pmod{3}$. We can substitute 2 for $A+B$, so $2+C\equiv 0 \pmod{3}$, and therefore $C\equiv 1\pmod{3}$. This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is $\boxed{\textbf{(A) }1}$.

Solution

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Another solution is using the divisbility rule of 3. Just add the digits of each number together.

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