A string of 2015 digits starts with a "4". Any number formed by two consecutive digits is divisible by either 19 or 23.
What is the last digit in this sequence?
What if the sequence had 2016 digits?
Following 46 only 69 is possible- later there are two possibilities:-
a: 4695769576957...6957 etc
b. 469238 .ended no 2digit multiple of 23 or 19 starts with 8
in case a we have 1 digit (4 ) followed 503 *( 6957) and thenon 2014 th place digit 6, on 2015 th place 9, answ:9
in case b we have 1 digit (4 ) followed by 503 * (6957) and then on 2014 th place 6, on 2015 th place 9, and the 2016th could be either 5 or 2
On 2015th place, no matter what only 9 could be the answer
On 2016th place, it could be either 5 or 2.<p>
solution
