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BALTIC-WAY 2017 Problem 19
For an integer n ≥ 1 n \geq 1 n ≥ 1 let a ( n ) a(n) a ( n ) denote the total number of carries which arise when adding 2017 and n ⋅ 2017 n \cdot 2017 n ⋅ 2017 . The first few values are given by a ( 1 ) = 1 , a ( 2 ) = 1 , a ( 3 ) = 0 a(1)=1, a(2)=1, a(3)=0 a ( 1 ) = 1 , a ( 2 ) = 1 , a ( 3 ) = 0 , which can be seen from the following:
001
001
000
2017
4034
6051
+2017
+2017
+2017
= 4034 =4034 = 4034
= 6051 =6051 = 6051
= 8068 =8068 = 8068
Prove that
a ( 1 ) + a ( 2 ) + … + a ( 10 2017 − 2 ) + a ( 10 2017 − 1 ) = 10 ⋅ 10 2017 − 1 9 a(1)+a(2)+\ldots+a\left(10^{2017}-2\right)+a\left(10^{2017}-1\right)=10 \cdot \frac{10^{2017}-1}{9} a ( 1 ) + a ( 2 ) + … + a ( 1 0 2017 − 2 ) + a ( 1 0 2017 − 1 ) = 10 ⋅ 9 1 0 2017 − 1
Dataset text NuminaMath olympiads . Quoted with source.
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10 \cdot \frac{10^{2017}-1}{9}
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