In a sequence , we have and .
After writing 2017 terms, how many times does the digit 8 appear in the units digit?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
a 1 = 3
a 2 = 2 × a 1 + 2 = 8
a 3 = 2 × 8 + 2 = 1 6 + 2 = 1 8
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 2 0 1 7 terms and only one of them ends in a digit other than 8 , there are total of 2 0 1 6 terms ending in 8 .