Difference between revisions of "2019 AIME I Problems/Problem 1"

Revision as of 08:44, 20 September 2019

Problem 1

Consider the integer $N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.$ Find the sum of the digits of $N$ .

Solution

Let's express the number in terms of $10^n$ . We can obtain $(10-1)+(10^2-1)+(10^3-1)+\cdots+(10^{321}-1)$ . By the commutative and associative property, we can group it into $(10+10^2+10^3+\cdots+10^{321})-321$ . We know the former will yield $1111....10$ , so we only have to figure out what the last few digits are. There are currently $321$ 1's. We know the last 4 digits are 1110, and that the others will not be affected if we subtract $321$ . If we do so, we get that $1110-321=789$ . This method will remove 3 1's, and add a 7, 8 and 9. Therefore, the sum of the digits is $(321-3)+7+8+9=\boxed{342}$ .

-eric2020

Solution 2

We can see that $9=9$ , $9+99=108$ , $9+99+999=1107$ , all the way to ten nines when we have $11111111100$ . Then, when we add more nines, we repeat the same process, and quickly get that the sum of digits is $\boxed{342}$ since we have to add $9\lfloor \log 321 \rfloor$ to the sum of digits, which is $9\lceil \frac{321}9 \rceil$ .

Video Solution

https://www.youtube.com/watch?v=JFHjpxoYLDk

@@ Line 7: / Line 7: @@
 -eric2020
+==Solution 2==
+We can see that <math>9=9</math>, <math>9+99=108</math>, <math>9+99+999=1107</math>, all the way to ten nines when we have <math>11111111100</math>. Then, when we add more nines, we repeat the same process, and quickly get that the sum of digits is <math>\boxed{342}</math> since we have to add <math>9\lfloor \log 321 \rfloor</math> to the sum of digits, which is <math>9\lceil \frac{321}9 \rceil</math>.
 ==Video Solution==

2019 AIME I (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search