Year in a sequence (2)

Algebra Level 3

In a sequence $Q$ , we have $a_1=3$ and $a_n= 2a_{n-1}+2$ .

After writing 2017 terms, how many times does the digit 8 appear in the units digit?

2019 2017 1008 1009 2016 2015

2 solutions

Marta Reece
Jul 5, 2017

$a_1=3$

$a_2=2\times a_1+2=8$

$a_3=2\times8+2=16+2=18$

The units digit after this continues to be dependent only on the units digit of the previous term, not on any of the higher order digits.

Therefore the calculation used for $a_3$ is applicable for all $a_n$ with $n>2$ as all of these terms are calculated based on a terms ending in $8$ and all in turn end in $8$ .

Since there are $2017$ terms and only one of them ends in a digit other than $8$ , there are total of $\boxed{2016}$ terms ending in $8$ .

Thank you.

Hana Wehbi - 3 years, 11 months ago

Hana Wehbi
Jul 5, 2017

By writing few terms, we notice all the terms have $8$ in the units digit except for $a_1$ , thus, the answer is $2016$

How do you know that ALL the terms have 8 in the unit digits? Did you do proof by exhaustion?

Pi Han Goh - 3 years, 11 months ago

Actually yes. But notice that every term has $8$ in the units digit, the unit digit of the preceding term which is $8$ is going to be multiplied by $2$ plus $2$ which is $8\times 2 +2=18$ which in turn we get $8$ again as the unit digit for the following term.

Hana Wehbi - 3 years, 11 months ago

Year in a sequence (2)

2 solutions

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