2016 AMC 8 Problems/Problem 24
Contents
Problem 24
The digits , , , , and are each used once to write a five-digit number . The three-digit number is divisible by , the three-digit number is divisible by , and the three-digit number is divisible by . What is ?
Solutions
Solution 1 (Modular Arithmetic)
We see that since is divisible by , must equal either or , but it cannot equal , so . We notice that since must be even, must be either or . However, when , we see that , which cannot happen because and are already used up; so . This gives , meaning . Now, we see that could be either or , but is not divisible by , but is. This means that and .
Solution 2
We know that out of is divisible by . Therefore is obviously 5 because is divisible by 5. So we now have as our number. Next, let's move on to the second piece of information that was given to us. RST is divisible by 3. So, according to the divisibility by 3 rule, the sum of has to be a multiple of 3. The only 2 big enough are 9 and 12 and since 5 is already given. The possible sums of are 4 and 7. So, the possible values for are 1,3,4,3 and the possible values of are 3,1,3,4. So, using this we can move on to the fact that is divisible by 4. So, using that we know that has to be even so 4 is the only possible value for . Using that we also know that 3 is the only possible value for 3. So, we have = so the possible values are 1 and 2 for and . Using the divisibility rule of 4 we know that has to be divisible by 4. So, either 14 or 24 are the possibilities, and 24 is divisible by 4. So the only value left for is 1. .
Solution 3 (Divisibility Rules)
We know that is divisible by , so would be either or . However, is not a choice, so . Also, is divisible by , so this means that is , , , or . If , then has to be or ( is divisible by ), but both are taken. So, . must equal or , but because , . This leaves
~MrThinker
Video Solution
https://youtu.be/WJ0Hodj0h2o - Happytwin
https://youtu.be/6xNkyDgIhEE?t=2905
~savannahsolver
See Also
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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.