Polynomials & Real Numbers: Find Remainder & No. of Divisors

[HallsofIvy]

Now you are just guessing. The critical point is that $10^{998}= (5^{998})(2^{998})$. That means that $5^n$ is a factor for every n from 0 to 998. But it also means that $2^n$, for n= 0 to 99 is also a factor. That's 2(999) right there, and and we haven't even started combining "2"s and "5"s.

One way to look at it is this: there are 999 ways to find factors using only "5"s and there are 999 ways using only "2"s. And the "fundamental theorem of counting" says that the if there are m ways of doing one thing and n ways of doing another (independently of the first) then there are mn ways of doing both.

 

[Curious3141]

Now you are just guessing. The critical point is that $10^{998}= (5^{998})(2^{998})$. That means that $5^n$ is a factor for every n from 0 to 998. But it also means that $2^n$, for n= 0 to 99 is also a factor. That's 2(999) right there, and and we haven't even started combining "2"s and "5"s.

Yes, but everything from $5^0$ to $5^{998}$ are also factors of $10^{998}$ and are therefore excluded from consideration. Ditto for powers of 2. Every number that should be considered should have as a factor, either $2^{999}$ or $5^{999}$ or both (the only one that has both is $10^{999}$).

 

[sambarbarian]

i don't quite understand

 

[Curious3141]

i don't quite understand

OK, start by considering the numbers with $2^{999}$ as a factor. Now, everything from $2^{999}.5^0$ to $2^{999}.5^{998}$ divides $10^{999}$ but not $10^{998}$. Agreed? So how many is that?

Now consider numbers with $5^{999}$ as a factor in a similar way. How many do we have here?

Finally, don't forget $2^{999}.5^{999} = 10^{999}$ itself, which I avoided including till the last step so that we don't double-count it.

 

[sambarbarian]

OK, start by considering the numbers with $2^{999}$ as a factor. Now, everything from $2^{999}.5^0$ to $2^{999}.5^{998}$ divides $10^{999}$ but not $10^{998}$. Agreed? So how many is that?

2^999 . 5^0 is equal to 2^999 . 5^1 ,,, (998 - 1) = 997 ? i am not sure

 

[Curious3141]

2^999 . 5^0 is equal to 2^999 . 5^1 ,,, (998 - 1) = 997 ? i am not sure

You're not thinking clearly, and what you wrote makes no sense.

$2^{999}.5^0 = 2^{999}$ because any nonzero number raised to the power zero is 1.

$2^{999}$ is still a number you have to include in the count because it divides $10^{999}$ but not $10^{998}$.

I asked you to count the numbers from $2^{999}.5^0$ to $2^{999}.5^{998}$ going in consecutive exponents of 5. This is equivalent to counting the integers from 0 to 998 inclusive. You replied with '997'. Sure of this?

How many integers are there from 0 to 1 inclusive? 0 to 10? 0 to 998?