+0  
 
+1
46
4
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Let \(a_n\) be the integer obtained by writing all the integers from 1 to \(n\) from left to right. For example, \(a_3 = 123\) and \(a_{11} = 1234567891011\). Compute the remainder when \(a_{44}\) is divided by 45.

 Jan 4, 2019
 #1
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0

a(44) mod 45 = 9

 Jan 4, 2019
 #2
avatar+95179 
+1

How do you know that guest?

Melody  Jan 5, 2019
 #3
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+3

By direct division:
(1234567891011121314151617181920212223242526272829303132333435363738394041424344) / 45 =
2743484 2022469362 5367026040 4267138273 8722806062 8734029407 4300808307 5423142763.2
0.2 x 45 = 9 - The remainder.

Guest Jan 5, 2019
 #4
avatar+51 
0

Hmm.

I have a very similar question as a matter of fact.

 

The only difference is that I am trying to compute the remainder when a_2017 is divided by 45.

 

Any thoughts?

 I couldn't find where the 44 was implemented in guest's answers, otherwise I would have attempted to solve this problem by plugging in 2017 in to his/her's method. 

It is clearly not viable to type all the first 2017 numbers into a calculator :D

 Jan 5, 2019
edited by sudsw12  Jan 5, 2019

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