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# How many?

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

Note by Dev Sharma
1 year, 7 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 $$10^{998}.$$ · 1 year, 7 months ago

Thanks sir.

You are magician. · 1 year, 7 months ago

A Mathemagician :) · 1 year, 7 months ago

Sir, I have one doubt. I think we should have to multiply 1000 by 1000 · 1 year, 7 months ago

In which place are you referring? · 1 year, 7 months ago

in above solution, sir has added 1000 and 1000. But I think that they should be multiply. · 1 year, 7 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. :) · 1 year, 7 months ago

Thanks. · 1 year, 7 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 $$1000+1000-1=1999$$. · 1 year, 7 months ago

Thanks sir · 1 year, 7 months ago

Haha. You're welcome. :) · 1 year, 7 months ago

your a mathemagician · 1 year, 7 months ago

Sir, I have one doubt.

I think we should have to multiply 1000 by 1000 · 1 year, 7 months ago

1999 · 1 year, 6 months ago

Hm, instead of constructively counting, I wonder if we can find a one to one correspondence. · 1 year, 7 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

$$225*10^{997} - 2^{997} - 5^{999}.$$ · 1 year, 7 months ago