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# arithmetic series

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What is the remainder when the sum 1 + 7 + 13 + 19 + ... + 253 + 259 + 265 + 271 is divided by 6?

Mar 16, 2021

#1
+507
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you can notice that you are adding 6 more to the previous number, starting with 1. if we divide each of these numbers by 6, we will always get a remainder of 1. we can subtract 1 from the last number and divide it by 6, to see how many numbers in total are in the sequence. we get (271-1)/6=270/6=45.

now that we know there are 45 digits and each of the remainders when divided by 6 is 1, we know that the remainder of the sum divided by 6 is 45/6=7 3/2=\$\boxed3\$

Mar 16, 2021
#2
+121004
+1

Terms in the  series =     (271  - 1)   /  6     +    1 = 270 / 6  +  1  =   45 + 1  =     46 terms

Each  term can be expressed  as    ( 1 + 6n )      where n = 0,1,2,3.....43, 44,45

And each term  divided by 6 leaves a remainder of 1

So   summing all the remainders  we  get   46

And  46 /  6  =   7 * 6  +  4   =    remainder  of  4

Mar 16, 2021
edited by CPhill  Mar 16, 2021