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

Revision as of 02:06, 24 December 2022

Problem 21

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 $C$ ?

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

Video Solution for Problems 21-25

https://www.youtube.com/watch?v=6S0u_fDjSxc

Video Solution

https://youtu.be/6xNkyDgIhEE?t=2593

https://youtu.be/7TOtBiod55Q ~savannahsolver

Solution 1

The number $\mod{3}$ is congruent to sum of a number's digits $\mod{3}$ is congruent to the number $\mod{3}$ . $74A52B1 \pmod{3}$ must be congruent to 0, since it is divisible by 3. Therefore, $7+4+A+5+2+B+1 \pmod{3}$ 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 2

Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. $7 + 4 + 5 + 2 + 1 = 19$ . To be a multiple of $3$ , $A + B$ has to be either $2$ or $5$ or $8$ ... and so on. We add up the numerical digits in the second number; $3 + 2 + 6 + 4 = 15$ . We then add two of the selected values, $5$ to $15$ , to get $20$ . We then see that C = $1, 4$ or $7, 10$ ... and so on, 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 $7$ ... 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 $1, 4,$ and $7$ . However, in the answer choices, there is no $7$ or $4$ or anything greater than $7$ , but there is a $1$ , so $\boxed{\textbf{(A) }1}$ is our answer.

@@ Line 12: / Line 12: @@
 ==Solution 1==
-The sum of a number's digits<math>\mod{3}</math> is congruent to the number<math>\mod{3}</math>. <math>74A52B1 \pmod{3}</math> must be congruent to 0, since it is divisible by 3. Therefore, <math>7+4+A+5+2+B+1 \pmod{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 number<math>\mod{3}</math> is congruent to sum of a number's digits<math>\mod{3}</math> is congruent to the number<math>\mod{3}</math>. <math>74A52B1 \pmod{3}</math> must be congruent to 0, since it is divisible by 3. Therefore, <math>7+4+A+5+2+B+1 \pmod{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>.
 ==Solution 2==

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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