Difference between revisions of "2014 AMC 8 Problems/Problem 21"
(→Solution 1) |
(→Video Solution for Problems 21-25) |
||
| Line 9: | Line 9: | ||
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. | 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 by OmegaLearn== | ==Video Solution by OmegaLearn== | ||
https://youtu.be/6xNkyDgIhEE?t=2593 | https://youtu.be/6xNkyDgIhEE?t=2593 | ||
Revision as of 08:51, 23 July 2024
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.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
