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 \(10^{998}.\)
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Brian Charlesworth
·
1 year, 4 months ago

@Daniel Liu
–
in above solution, sir has added 1000 and 1000. But I think that they should be multiply.
–
Dev Sharma
·
1 year, 4 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. :)
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Brian Charlesworth
·
1 year, 4 months ago

We have that \((m,n)=(999,0), (999,1), \ldots (999,999), \ldots (1, 999), (0, 999)\) are solutions. To count this, we compute \(1000+1000-1=1999\).
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Daniel Liu
·
1 year, 4 months ago

Hm, instead of constructively counting, I wonder if we can find a one to one correspondence.
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Alan Yan
·
1 year, 4 months ago

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@Alan Yan
–
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

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TopNewestSince \(10^{999} = 2^{999}5^{999}\) 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 · 1 year, 4 months ago

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You are magician. – Dev Sharma · 1 year, 4 months ago

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– Daniel Liu · 1 year, 4 months ago

A Mathemagician :)Log in to reply

– Dev Sharma · 1 year, 4 months ago

Sir, I have one doubt. I think we should have to multiply 1000 by 1000Log in to reply

– Daniel Liu · 1 year, 4 months ago

In which place are you referring?Log in to reply

– Dev Sharma · 1 year, 4 months ago

in above solution, sir has added 1000 and 1000. But I think that they should be multiply.Log in to reply

– Brian Charlesworth · 1 year, 4 months ago

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. :)Log in to reply

– Dev Sharma · 1 year, 4 months ago

Thanks.Log in to reply

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 · 1 year, 4 months ago

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– Dev Sharma · 1 year, 4 months ago

Thanks sirLog in to reply

– Brian Charlesworth · 1 year, 4 months ago

Haha. You're welcome. :)Log in to reply

– Ishita .S · 1 year, 4 months ago

your a mathemagicianLog in to reply

I think we should have to multiply 1000 by 1000 – Dev Sharma · 1 year, 4 months ago

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1999 – Amenreet Singh Sodhi · 1 year, 4 months ago

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Hm, instead of constructively counting, I wonder if we can find a one to one correspondence. – Alan Yan · 1 year, 4 months ago

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

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 · 1 year, 4 months ago

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