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sum of nine consecutive quotients of a_{n}= 1111...111 is a multiple of 9

Source: Greece JBMO TST 2014 p3

April 29, 2019

number theoryconsecutiveSummultiple

Problem Statement

Give are the integers

a_{1}=11 , a_{2}=1111, a_{3}=111111, ... , a_{n}= 1111...111

( with

2n

digits) with

n > 8

. Let

q_{i}= \frac{a_{i}}{11} , i= 1,2,3, ... , n

the remainder of the division of

a_{i}

11

. Prove that the sum of nine consecutive quotients:

s_{i}=q_{i}+q_{i+1}+q_{i+2}+ ... +q_{i+8}

is a multiple of

9

for any

i= 1,2,3, ... , (n-8)