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Find the number of positive integers which divide \(10^{999}\) but not \(10^{998}\)

Note by Dev Sharma
2 years, 3 months ago

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Since \(10^{999} = 2^{999}5^{999}\) and \(10^{998} = 2^{998}5^{998},\) all numbers of the form \(2^{m}5^{n}, 0 \le m,n \le 999\) where at least one of \(m,n\) equals \(999\) will divide \(10^{999}\) but not \(10^{998}.\)

So with \(m = 999\) we can have \(n\) taking on \(1000\) integer values from \(0\) to \(999.\) The same goes for \(n = 999\) with \(m\) taking on \(1000\) values, but to avoid double-counting we must remember to count \(m = n = 999\) only once, resulting in a total of \(1000 + 1000 - 1 = 1999\) positive divisors of \(10^{999}\) which do not divide \(10^{998}.\)

Brian Charlesworth - 2 years, 3 months ago

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Thanks sir.

You are magician.

Dev Sharma - 2 years, 3 months ago

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A Mathemagician :)

Daniel Liu - 2 years, 3 months ago

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@Daniel Liu Sir, I have one doubt. I think we should have to multiply 1000 by 1000

Dev Sharma - 2 years, 3 months ago

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@Dev Sharma In which place are you referring?

Daniel Liu - 2 years, 3 months ago

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@Daniel Liu in above solution, sir has added 1000 and 1000. But I think that they should be multiply.

Dev Sharma - 2 years, 3 months ago

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@Dev Sharma For this problem addition is the correct operation. Note that \(2^{999}5^{999}\) has \((999 + 1)(999 + 1) = 10^{6}\) positive divisors; perhaps that was what you were thinking about. :)

Brian Charlesworth - 2 years, 3 months ago

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@Brian Charlesworth Thanks.

Dev Sharma - 2 years, 3 months ago

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@Dev Sharma It's supposed to be add.

We have that \((m,n)=(999,0), (999,1), \ldots (999,999), \ldots (1, 999), (0, 999)\) are solutions. To count this, we compute \(1000+1000-1=1999\).

Daniel Liu - 2 years, 3 months ago

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@Daniel Liu Thanks sir

Dev Sharma - 2 years, 3 months ago

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Haha. You're welcome. :)

Brian Charlesworth - 2 years, 3 months ago

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your a mathemagician

Ishita .S - 2 years, 3 months ago

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Sir, I have one doubt.

I think we should have to multiply 1000 by 1000

Dev Sharma - 2 years, 3 months ago

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1999

Amenreet Singh Sodhi - 2 years, 3 months ago

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Hm, instead of constructively counting, I wonder if we can find a one to one correspondence.

Alan Yan - 2 years, 3 months ago

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Come to think of it, it's probably easiest to just subtract the number of positive divisors of \(10^{998} = 2^{998}5^{998}\) from the number of positive divisors of \(10^{999} = 2^{999}5^{999}.\) This gives us an answer of

\((999 + 1)(999 + 1) - (998 + 1)(998 + 1) = 1000^{2} - 999^{2} = (1000 - 999)(1000 + 999) = 1999.\)

A slightly more involved question would be to find the sum of these \(1999\) divisors. I get an answer of

\(225*10^{997} - 2^{997} - 5^{999}.\)

Brian Charlesworth - 2 years, 3 months ago

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