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 Entertain Every Value? (Posted on 2014-04-29)
The respective number of digits of 4n and 25n - when added together is odd, whenever n is a positive integer.

Does this assertion hold for every value of n?

If so, prove it.
If not, provide a counter example.

 No Solution Yet Submitted by K Sengupta No Rating

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 proof (acceptable?) | Comment 1 of 2
The number of digits of a number is given by [log(x)]+1
where [ ] denotes the greatest integer and the logarithm is base 10.

So we are trying to show [log(4n)]+1 +[log(25n)]+1 is odd
which is equivalent to showing [log(4n)]+[log(25n)] is odd

[log(4n)]+[log(25n)] = [n*log(4)]+[n*log(25)]
= [n*(2-log(25))]+[n*log(25)]
= [2n - n*log(25)]+[n*log(25)]

What I now need to show is for any integer a and non-integer b
[a-b]+[b] = a-1
which it is because
[a-b]=a+[-b]
and
[-b]+[b]=-1

So the above expression becomes 2n-1

 Posted by Jer on 2014-04-29 14:10:04

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