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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Feb 2008 22:26:38 IST
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Yes it can solved by mathematical induction.
Another method : if last digit of n2+n+1 is either 0 or 5 then only it is divisible by 5.
When last digit of n is 0, last digit of n2+n+1 is 1.
When last digit of n is 1, last digit of n2+n+1 is 3.
When last digit of n is 2, last digit of n2+n+1 is 7.
When last digit of n is 3, last digit of n2+n+1 is 3.
When last digit of n is 4, last digit of n2+n+1 is 1.
When last digit of n is 5, last digit of n2+n+1 is 1.
When last digit of n is 6, last digit of n2+n+1 is 3.
When last digit of n is 7, last digit of n2+n+1 is 7.
When last digit of n is 8, last digit of n2+n+1 is 3.
When last digit of n is 9, last digit of n2+n+1 is 1.
In any of the cases last digit is neither 0 nor 5, hence n2+n+1 is not divisible by 5.
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Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur
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this reply: 15 points
(with 3 
in 3 votes ) [?]
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