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

Latest revision as of 20:26, 17 November 2024

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 $C$ ?

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

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.

Concept Solution

for a number to be divisible by 3, the sum of its digits must be divisible by 3. In the number 74A52B1, 7+4+A+5+2+B+1 must be divisible by 3. We get 19+A+B. 21 is the closest multiple of 3 meaning that A=2, B=0, order doesn't matter. Now we plug in those values for 326AB4C. It will be 3+2+6+2+0+4+C to get 17+C. The closest multiple of 3 is 18. So that means 17+C=18. Solving for C, our answer is $\boxed{\textbf{(A) }1}$ -TheNerdWhoIsNerdy.

Video Solution by Pi Academy (Fast and Easy ⚡️🚀)

https://youtu.be/t-SjP-gqw20?si=y6ECC_zEYV5OcxY2

Video Solution by OmegaLearn

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

Video Solution

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

@@ Line 1: / Line 1: @@
-==Problem==
+==Problem 21==
 The <math>7</math>-digit numbers <math>\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}</math> and <math>\underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C}</math> are each multiples of <math>3</math>. Which of the following could be the value of <math>C</math>?
 <math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8</math>
-==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>\textbf{(A) }1</math>.
+==Solution 2==
+Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. <math>7 + 4 + 5 + 2 + 1 = 19</math>. To be a multiple of <math>3</math>, <math>A + B</math> has to be either <math>2</math> or <math>5</math> or <math>8</math>... and so on. We add up the numerical digits in the second number; <math>3 + 2 + 6 + 4 = 15</math>. We then add two of the selected values, <math>5</math> to <math>15</math>, to get <math>20</math>. We then see that C = <math>1, 4</math> or <math>7, 10</math>... and so on, otherwise the number will not be divisible by three. We then add <math>8</math> to <math>15</math>, to get <math>23</math>, which shows us that C = <math>1</math> or <math>4</math> or <math>7</math>... 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 <math>1, 4,</math> and <math>7</math>. However, in the answer choices, there is no <math>7</math> or <math>4</math> or anything greater than <math>7</math>, but there is a <math>1</math>, so  <math>\boxed{\textbf{(A) }1}</math> is our answer.
+== Concept Solution==
+for a number to be divisible by 3, the sum of its digits must be divisible by 3. In the number 74A52B1, 7+4+A+5+2+B+1 must be divisible by 3. We get 19+A+B. 21 is the closest multiple of 3 meaning that A=2, B=0, order doesn't matter. Now we plug in those values for 326AB4C. It will be 3+2+6+2+0+4+C to get 17+C. The closest multiple of 3 is 18. So that means 17+C=18. Solving for C, our answer is <math>\boxed{\textbf{(A) }1}</math>-TheNerdWhoIsNerdy.
+== Video Solution by Pi Academy (Fast and Easy ⚡️🚀) ==
+https://youtu.be/t-SjP-gqw20?si=y6ECC_zEYV5OcxY2
+== Video Solution by OmegaLearn==
+https://youtu.be/6xNkyDgIhEE?t=2593
+==Video Solution==
+https://youtu.be/7TOtBiod55Q ~savannahsolver
 ==See Also==
-{{AMC8 box|month=2014|num-b=20|num-a=22}}
+{{AMC8 box|year=2014|num-b=20|num-a=22}}
 {{MAA Notice}}

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