SOLUTION: A list of whole numbers from 1 to 2015 is written on a sheet of paper. All the multiples of 5 are then struck off from the list. What is the last digit of the product of the remain

Algebra ->  Divisibility and Prime Numbers -> SOLUTION: A list of whole numbers from 1 to 2015 is written on a sheet of paper. All the multiples of 5 are then struck off from the list. What is the last digit of the product of the remain      Log On


   



Question 1162771: A list of whole numbers from 1 to 2015 is written on a sheet of paper. All the multiples of 5 are then struck off from the list. What is the last digit of the product of the remaining numbers?
Answer by ikleyn(52921) About Me  (Show Source):
You can put this solution on YOUR website!
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To get the answer, we should look and trace the last digit, only.


Regarding last digits, for our first numbers 1, 2, 3, 4, highlight%28cross%285%29%29, 6, 7, 8, 9, highlight%28cross%2810%29%29 the last digit of the product is


    1*2*3*4*6*7*8*9 = 6 (mod10)   <<<---===  I wrote the LAST digit only (!!!)


Now, these factors will be cyclically repeated for every 10 numbers (with excluded 5 and 0).


Therefore, the last digit of the product, created by every 10 numbers, is 6.


    6*6 = 36 = 6 (mod10),


therefore, the last digit of the product of all these factors from 1 to 2010 (with excluded multiples of 5) is 6.


Now, to answer the final question, we should calculate

  
    6 * 11*12*13*14.


or, tracing the last digit only,  the product


    6 * 1*2*3*4 = 4  (mod10).


ANSWER.  The last digit of the number under the question is 4.

Solved.