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Consider the sequence a_1 = 1, a_2 = 13, a_3 = 135,... of positive integers. The k^th term a_k is defined by appending the digits of the k^th odd integer to the preceding term a_k-1. For example, since the 5^th term is a_5 = 13579, then the 6th term is a_6 = 1357911. Note that a_3 is the first term of the sequence divisible by 9. What is the value of m such that a_m is the 23^rd term of the sequence that is divisible by 9.

 

Thank you so much if you can help.

 Oct 26, 2019
 #1
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If I understand your sequence, it should go like this:


1, 13, 135, 1357, 13579, 1357911, 135971113.....etc.  Note: Every 3rd term is divisible by 9. That is: 3rd, 6th, 9th, 12th, 15th, 18th, 21st and 24th...etc. The 23rd term that is divisible by 9 is actually the 69th term of the sequence, since the first term that is divisible by 9 is the 3rd term, and they go up as 3n, so the 23rd term is:23 x 3 =69th term of the sequence, and it is this long!.

 

 

=13579111315171921232527293133353739414345474951535557596163656769717375777981838587899193959799101103105107109111113115117119121123125127129131133135137. It is 152 digits long and sums up to:612 which is divisible by 9.Or = 612 / 9 = 68

 

So, a(m) = 69th term

 Oct 27, 2019
 #2
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Thank You So Much

Guest Oct 27, 2019

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