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Topic: Consider The Minimum, Get 98 Zeroes (Read 279 times) |
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K Sengupta
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Consider The Minimum, Get 98 Zeroes
« on: Jul 24th, 2007, 7:19am » |
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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 aa*bb are all zeroes.
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SMQ
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Re: Consider The Minimum, Get 98 Zeroes
« Reply #1 on: Jul 24th, 2007, 1:47pm » |
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At least the last 98, or only the last 98? In the former case, I don't see a way to beat the trivial solution of 1, 100; in the latter case, the family 75+150n, 98 (where n = 0 gives the obvious minimum) seem to be the only solutions.
--SMQ
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Eigenray
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Re: Consider The Minimum, Get 98 Zeroes
« Reply #2 on: Jul 24th, 2007, 2:58pm » |
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SMQ, I get the same minimum, but your general solution only gives ~ 1/15 of the solutions.
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The number of times 5 divides n=aabb is v5(n) = a*v5(a)+b*v5(b), which is evidently divisible by 5. Since v5(n)  98, in fact v5(n) 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) 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=2rm, with m odd, this gives r*2rm=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=2rm as before gives r+v2(r)=1, so r=1, and b=98.
Now a*v5(a) 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 0}. |
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SMQ
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Re: Consider The Minimum, Get 98 Zeroes
« Reply #3 on: Jul 24th, 2007, 4:22pm » |
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Yeah, I realized after I posted it and before I could get back to a computer that 98, any odd multiple of 5 > 98 would work.
--SMQ
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| « Last Edit: Jul 24th, 2007, 4:22pm by SMQ » |
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