2017 AMC 12A Problems/Problem 18
Contents
Problem
Let equal the sum of the digits of positive integer . For example, . For a particular positive integer , . Which of the following could be the value of ?
Solution 1
Note that , so . So, since , we have that . The only one of the answer choices is .
Solution 2
One possible value of would be , but this is not any of the choices. Therefore, we know that ends in , and after adding , the last digit carries over, turning the last digit into . If the next digit is also a , this process repeats until we get to a non- digit. By the end, the sum of digits would decrease by multiplied by the number of carry-overs but increase by as a result of the final carrying over. Therefore, the result must be less than original value of , , where is a positive integer. The only choice that satisfies this condition is , since . The answer is .
Solution 3
Another way to solve this is to realize that if you continuously add the digits of the number , we get . Adding one to that, we get . So, if we assess each option to see which one attains , we would discover that satisfies the requirement, because . . The answer is .
Solution 4(Similar to Solution 1)
Note that a lot of numbers can have a sum of , but what we use wishful thinking and want is some simple number where it is easy to compute the sum of the digits of . This number would consists of basically all digits , since when you add a lot of stuff will cancel out and end up at (ex: ). We see that the maximum number of s that can be in is and we are left with a remainder of , so is in the form . If we add to this number we will get so this the sum of the digits of is congruent to . The only answer choice that is equivalent to is , so our answer is -srisainandan6
Video Solution by OmegaLearn
https://youtu.be/zfChnbMGLVQ?t=3996
~ pi_is_3.14
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 AMC 10 Problems and Solutions |
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 AMC 12 Problems and Solutions |
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