Faster Method than Brute Force (Find sum of consecutive terms)

Asked Nov 24 '17 at 23:05

Active Nov 24 '17 at 23:05

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Let $a_n$ be the number obtained by writing the integers 1 to $n$ from left to right. Therefore, $a_4 = 1234$ and a_12 = 123456789101112. For $1 \le k \le 100$, how many $a_k$ are divisible by 9?

I know that to find this amount, you need to find the sum of the digits, and then see if it is a factor of 9. However, I want to know a method other than using brute force. Can someone guide me through a slicker solution?

asked Nov 24 '17 at 23:05

A Piercing Arrow

Here is something to think about. For $a_1$ through $a_9$, which ones are multiples of $9$? How about for $a_{10}$ through $a_{18}$? Can you see a pattern emerging? – John Brevik Nov 24 '17 at 23:11
Factors of 9...I entered 11, but it didn't work. – A Piercing Arrow Nov 24 '17 at 23:12
I don't follow. – John Brevik Nov 24 '17 at 23:13
Oh...From 1 - 9 , there is 1 that works. From 10-19, there are 2. – A Piercing Arrow Nov 24 '17 at 23:15
Are you sure? Recheck your work. Which exact ones? – John Brevik Nov 24 '17 at 23:17

Faster Method than Brute Force (Find sum of consecutive terms)

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