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        riddles >> medium >> Consider The Minimum, Get 98 Zeroes
        
(Message started by: K Sengupta on Jul 24^th, 2007, 7:19am)

Title: Consider The Minimum, Get 98 Zeroes
Post by K Sengupta on Jul 24^th, 2007, 7:19am

Analytically determine the positive integers a and b , so that ab is minimum with the proviso that the last 98 digits in the decimal expansion of a^a*b^b are all zeroes.

Title: Re: Consider The Minimum, Get 98 Zeroes
Post by SMQ on Jul 24^th, 2007, 1:47pm

At least the last 98, or only the last 98? In the former case, [hide]I don't see a way to beat the trivial solution of 1, 100[/hide]; in the latter case, [hide]the family 75+150n, 98 (where n = 0 gives the obvious minimum) seem to be the only solutions.[/hide]

--SMQ

Title: Re: Consider The Minimum, Get 98 Zeroes
Post by Eigenray on Jul 24^th, 2007, 2:58pm

SMQ, I get the same minimum, but your general solution only gives ~ 1/15 of the solutions.

[hideb]
The number of times 5 divides n=a^ab^b is v₅(n) = a*v₅(a)+b*v₅(b), which is evidently divisible by 5. Since v₅(n) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 98, in fact v₅(n) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 100. If only the last 98 digits of n are 0, then we must therefore have v₂(n) = 98. Therefore only one of a,b can be divisible by 5; WLOG say 5|a.

If a is even, then since v₂(n)=98, we must have a < 100. But a*v₅(a) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 100, so a must be divisible by 25. The only possibility then is a=50, and so we need b*v₂(b) = 48. But if b=2^rm, with m odd, this gives r*2^rm=48=2⁴*3. So we need r + v₂(r) = 4, and checking r=1,2,3,4 shows this is impossible. This contradiction shows that a is odd.

We therefore have b*v₂(b) = 98; setting b=2^rm as before gives r+v₂(r)=1, so r=1, and b=98.

Now a*v₅(a) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 100. If a < 100, then 25|a, and the only possibility is a=75. For a > 100, any odd number divisible by 5 will do. So the possibilities for a are: 75, or {105+10k, k http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 0}.[/hideb]

Title: Re: Consider The Minimum, Get 98 Zeroes
Post by SMQ on Jul 24^th, 2007, 4:22pm

Yeah, I realized after I posted it and before I could get back to a computer that [hide]98, any odd multiple of 5 > 98[/hide] would work.

--SMQ