# How many?

Find the number of positive integers which divide $10^{999}$ but not $10^{998}$ Note by Dev Sharma
5 years, 11 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 $10^{998}.$

- 5 years, 11 months ago

Thanks sir.

You are magician.

- 5 years, 11 months ago

A Mathemagician :)

- 5 years, 11 months ago

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

- 5 years, 11 months ago

In which place are you referring?

- 5 years, 11 months ago

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

- 5 years, 11 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 $1000+1000-1=1999$.

- 5 years, 11 months ago

Thanks sir

- 5 years, 11 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. :)

- 5 years, 11 months ago

Thanks.

- 5 years, 11 months ago

Haha. You're welcome. :)

- 5 years, 11 months ago

- 5 years, 11 months ago

Sir, I have one doubt.

I think we should have to multiply 1000 by 1000

- 5 years, 11 months ago

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

- 5 years, 11 months ago

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

$225*10^{997} - 2^{997} - 5^{999}.$

- 5 years, 11 months ago

1999

- 5 years, 10 months ago