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2015 digits (Posted on 2014-09-04) Difficulty: 2 of 5
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?

No Solution Yet Submitted by K Sengupta    
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solution | Comment 1 of 2

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>


  Posted by Ady TZIDON on 2014-09-04 10:18:02
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