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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≤n≤100, how many a_n are divisible by 9?

 Jul 14, 2019
 #1
avatar+8576 
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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 ≤ n ≤ 100, how many \(a_n\) are divisible by 9?

 

 

Hello xXxTenTacion!

 

\(checksum\\ \sum^i_{i=1}(n_i)=(1+i)\cdot \frac{i}{2}\)

1  1

12  3

123  6 

1234  10

12345  15

123456  21

1234567  28

                  36  45  55  66  78  91  105  121 ... 5050

12345...100  5050

 

Sorry. I have to give up.

sad 

 Jul 14, 2019
edited by asinus  Jul 14, 2019
edited by asinus  Jul 14, 2019
 #2
avatar
+1

There are 22 "a_n" that are divisible by 9:

 

a_8, 9, 17, 18, 26, 27, 35, 36, 44, 45, 53, 54, 62, 63, 71, 72, 80, 81, 89, 90, 98, 99.

 Jul 14, 2019

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