How many?

Find the number of positive integers which divide 1099910^{999} but not 1099810^{998}

Note by Dev Sharma
4 years, 1 month ago

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Since 10999=2999599910^{999} = 2^{999}5^{999} and 10998=29985998,10^{998} = 2^{998}5^{998}, all numbers of the form 2m5n,0m,n9992^{m}5^{n}, 0 \le m,n \le 999 where at least one of m,nm,n equals 999999 will divide 1099910^{999} but not 10998.10^{998}.

So with m=999m = 999 we can have nn taking on 10001000 integer values from 00 to 999.999. The same goes for n=999n = 999 with mm taking on 10001000 values, but to avoid double-counting we must remember to count m=n=999m = n = 999 only once, resulting in a total of 1000+10001=19991000 + 1000 - 1 = 1999 positive divisors of 1099910^{999} which do not divide 10998.10^{998}.

Brian Charlesworth - 4 years, 1 month ago

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Thanks sir.

You are magician.

Dev Sharma - 4 years, 1 month ago

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A Mathemagician :)

Daniel Liu - 4 years, 1 month ago

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

Dev Sharma - 4 years, 1 month ago

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@Dev Sharma In which place are you referring?

Daniel Liu - 4 years, 1 month 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 - 4 years, 1 month ago

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@Dev Sharma It's supposed to be add.

We have that (m,n)=(999,0),(999,1),(999,999),(1,999),(0,999)(m,n)=(999,0), (999,1), \ldots (999,999), \ldots (1, 999), (0, 999) are solutions. To count this, we compute 1000+10001=19991000+1000-1=1999.

Daniel Liu - 4 years, 1 month ago

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@Daniel Liu Thanks sir

Dev Sharma - 4 years, 1 month ago

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@Dev Sharma For this problem addition is the correct operation. Note that 299959992^{999}5^{999} has (999+1)(999+1)=106(999 + 1)(999 + 1) = 10^{6} positive divisors; perhaps that was what you were thinking about. :)

Brian Charlesworth - 4 years, 1 month ago

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@Brian Charlesworth Thanks.

Dev Sharma - 4 years, 1 month ago

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Haha. You're welcome. :)

Brian Charlesworth - 4 years, 1 month ago

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your a mathemagician

Ishita .S - 4 years, 1 month ago

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Sir, I have one doubt.

I think we should have to multiply 1000 by 1000

Dev Sharma - 4 years, 1 month ago

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

Alan Yan - 4 years, 1 month ago

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Come to think of it, it's probably easiest to just subtract the number of positive divisors of 10998=2998599810^{998} = 2^{998}5^{998} from the number of positive divisors of 10999=29995999.10^{999} = 2^{999}5^{999}. This gives us an answer of

(999+1)(999+1)(998+1)(998+1)=100029992=(1000999)(1000+999)=1999.(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 19991999 divisors. I get an answer of

2251099729975999.225*10^{997} - 2^{997} - 5^{999}.

Brian Charlesworth - 4 years, 1 month ago

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1999

Amenreet Singh Sodhi - 4 years, 1 month ago

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