2019 AIME I Problems/Problem 1
Consider the integer Find the sum of the digits of .
Let's express the number in terms of . We can obtain . By the commutative and associative property, we can group it into . We know the former will yield , so we only have to figure out what the last few digits are. There are currently 1's. We know the last four digits are , and that the others will not be affected if we subtract . If we do so, we get that . This method will remove three 's, and add a , and . Therefore, the sum of the digits is .
A similar and simpler way to consider the initial manipulations is to observe that adding to each term results in . There are terms, so it becomes , where there are digits in . Then, subtract the you initially added.
We can see that , , , all the way to ten nines when we have . Then, when we add more nines, we repeat the same process, and quickly get that the sum of digits is since we have to add to the sum of digits, which is .
Solution 3 (Pattern)
Observe how adding results in the last term but with a concatenated in front and also a subtracted (, , , ). Then for any index of terms, , the sum is , where the first term is of length . Here, that is .
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