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 Ending in 5 (Posted on 2015-08-11)
N is a positive integer ending in 5 – that is, positive integers like 2015, 21335, 349705 etc.

Under what conditions do gcd(N, floor(N2/100)) exceed 1?

(For example, if N = 2015, then floor(N2/100) = 40602, so that:
gcd (N, floor(N2/100)) = 1, which contravenes the provisions of the problem.)

 No Solution Yet Submitted by K Sengupta No Rating

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 Possible solution Comment 1 of 1
All number ending in 5 can be written as one of the following:
50a+5
50a+15
50a+25
50a+35
50a+45 (=50a-5)

For the provisions of the problem to hold, n must be of the form 50a+5 or 50a-5
If n=50a+5=5(10a+1)
n²=2500a^2 + 500a + 25
floor(n²/100)=25a²+5a=5a(5a+1)
so the gcd is 5

If n=50a-5
n²=2500a^2 - 500a + 25
floor(n²/100)=25a²-5a
so the gcd is 5

Why can't other numbers work?
The others to consider are 50a+15, 50a+25, 50a+35
If n=50a+15=5(10a+3)
n^2=2500a²+1500a+225
floor(n²/100)=25a²+15a+2=(5a+1)(5a+2)
considering the factors mod 5 it is clear the gcd must be 1.

If n=50a+25=5(10a+5)=25(5a+1)
floor(n²/100)=(5a+2)(5a+3)
considering the factors mod 5 it is clear the gcd must be 1.

If n=50a+35=5(10a+7)
floor(n²/100)=(5a+3)(5a+4)
considering the factors mod 5 it is clear the gcd must be 1.

 Posted by Jer on 2015-08-11 11:34:13

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