Difference between revisions of "2014 AMC 8 Problems/Problem 21"
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==Solution 2== | ==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. In order 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. In order 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 <math>\boxed{\textbf{(A) }1}</math> is your answer. | + | 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>. In order 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. In order 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 your answer. |
==See Also== | ==See Also== | ||
Revision as of 15:25, 10 July 2021
Problem
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. 
. In order 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. In order 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 your 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
Another solution is using the divisbility rule of 3. Just add the digits of each number together.
