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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>
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==Solution 1==
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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>.
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==Solution 2==
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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.
  
 
==Video Solution for Problems 21-25==
 
==Video Solution for Problems 21-25==
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https://youtu.be/7TOtBiod55Q ~savannahsolver
 
https://youtu.be/7TOtBiod55Q ~savannahsolver
−
 
−
==Solution 1==
 
−
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==
 
−
 
−
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.
 
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==See Also==
 
==See Also==
 
{{AMC8 box|year=2014|num-b=20|num-a=22}}
 
{{AMC8 box|year=2014|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:10, 31 January 2023

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 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.

Video Solution for Problems 21-25

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

Video Solution by OmegaLearn

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

Video Solution

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

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

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