Difference between revisions of "2014 AMC 8 Problems/Problem 21"
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− | ==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>? | 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> | <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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==Video Solution== | ==Video Solution== | ||
https://youtu.be/6xNkyDgIhEE?t=2593 | https://youtu.be/6xNkyDgIhEE?t=2593 |
Revision as of 16:54, 5 January 2022
Problem 21
The -digit numbers and are each multiples of . Which of the following could be the value of ?
Video Solution
https://youtu.be/6xNkyDgIhEE?t=2593
Solution 1
The sum of a number's digits is congruent to the number. must be congruent to 0, since it is divisible by 3. Therefore, is also congruent to 0. , so . As we know, , so , and therefore . We can substitute 2 for , so , and therefore . This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is .
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.
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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