Difference between revisions of "2014 AMC 8 Problems/Problem 21"
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− | + | ==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>? | ||
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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> | ||
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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. | ||
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+ | == 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. | ||
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+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/6xNkyDgIhEE?t=2593 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/7TOtBiod55Q ~savannahsolver | ||
==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}} |
Latest revision as of 16:57, 15 September 2024
Contents
Problem 21
The -digit numbers and are each multiples of . Which of the following could be the value of ?
Solution 2
Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. . To be a multiple of , has to be either or or ... and so on. We add up the numerical digits in the second number; . We then add two of the selected values, to , to get . We then see that C = or ... and so on, otherwise the number will not be divisible by three. We then add to , to get , which shows us that C = or or ... 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 and . However, in the answer choices, there is no or or anything greater than , but there is a , so 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 -TheNerdWhoIsNerdy.
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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