Difference between revisions of "2017 AMC 12A Problems/Problem 18"
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==Solution 3== | ==Solution 3== | ||
Another way to solve this is to realize that if you continuously add the digits of the number <math>1274 (1 + 2 + 7 + 4 = 14, 1 + 4 = 5)</math>, we get <math>5</math>. Adding one to that, we get <math>6</math>. So, if we assess each option to see which one attains <math>6</math>, we would discover that <math>1239</math> satisfies the requirement, because <math>1 + 2 + 3 + 9 = 15</math>. <math>1 + 5 = 6</math>. The answer is <math>\boxed{D}</math>. | Another way to solve this is to realize that if you continuously add the digits of the number <math>1274 (1 + 2 + 7 + 4 = 14, 1 + 4 = 5)</math>, we get <math>5</math>. Adding one to that, we get <math>6</math>. So, if we assess each option to see which one attains <math>6</math>, we would discover that <math>1239</math> satisfies the requirement, because <math>1 + 2 + 3 + 9 = 15</math>. <math>1 + 5 = 6</math>. The answer is <math>\boxed{D}</math>. | ||
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+ | ==Solution 4(Similar to Solution 1)== | ||
+ | Note that a lot of numbers can have a sum of <math>1274</math>, but what we use wishful thinking and want is some simple number <math>n</math> where it is easy to compute the sum of the digits of <math>n+1</math>. This number would consists of basically all digits <math>9</math>, since when you add <math>1</math> a lot of stuff will cancel out and end up at <math>0</math>(ex: <math>399+1=400</math>). We see that the maximum number of <math>9</math>s that can be in <math>1274</math> is <math>141</math> and we are left with a remainder of <math>5</math>, so <math>n</math> is in the form <math>99...9599...9</math>. If we add <math>1</math> to this number we will get <math>99...9600...0</math> so this the sum of the digits of <math>n+1</math> is congruent to <math>6 \mod 9</math>. The only answer choice that is equivalent to <math>6 \mod 9</math> is <math>1239</math>, so our answer is <math>\boxed{D}</math> -srisainandan6 | ||
== See Also == | == See Also == |
Revision as of 17:03, 2 January 2021
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.