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- Last digits number theory. $7^{9999}$? - Mathematics Stack Exchange
If we want the last two digits, we note that $\phi (1000)=400$ So $$ 9999 = 9600 + 399$$ So $$ 7^ {9999} \equiv 7^ {399} \mod 1000 $$ Since $399$ is 1 less than $400$ we can calculate the answer easily
- Calculate $\binom {1000} {3}+\binom {1000} {8}+\binom {1000} {13 . . .
Hence , I am looking for helps to find a closed formula for the binomial expansion by simplifying $ (1+1)^ {1000}+w^2 (1+w)^ {1000}+w^4 (1+w^2)^ {1000}+w^6 (1+w^3)^ {1000}+w^8 (1+w^4)^ {1000}$ ADDENTUM: I want to reach an integer solution, as it is expected from this expression
- Which is bigger, $ \log_ {1000} 1001$ or $\log_ {999} 1000
It doesnt let me make it look correct, it's supposed to be log 1000 of 1001 and log 999 of 1000
- combinatorics - Determine the number of odd binomial coefficients in . . .
4 Determine the number of odd binomial coefficients in the expansion of $ (x+y)^ {1000}$ Hint: The number of odd coefficients in any finite binomial expansion is a power of $2$ Is there a way to prove this without using something like Lucas's theorem or any other non-trivial result?
- Keep rolling two dice until the cumulative sum hits 1000
Keep rolling two dice until the cumulative sum hits 1000 Ask Question Asked 2 years, 3 months ago Modified 2 years, 3 months ago
- Why is kg m³ to g cm³1 to 1000? - Mathematics Stack Exchange
I understand that changing the divisor multiplies the result by that, but why doesn't changing the numerator cancel that out? I found out somewhere else since posting, is there a way to delete this?
- definition - What is the smallest binary number of $4$ bit? Is it . . .
In pure math, the correct answer is $ (1000)_2$ Here's why Firstly, we have to understand that the leading zeros at any number system has no value likewise decimal Let's consider $2$ numbers One is $ (010)_2$ and another one is $ (010)_ {10}$ let's work with the $2$ nd number $ (010)_ {10}= (10)_ {10}$ We all agree that the smallest $2$ digit number is $10$ (decimal) Can't we say $010
- How much zeros has the number $1000!$ at the end?
1 If a number ends with n n zeros than it is divisible by 10n 10 n, that is 2n5n 2 n 5 n A factorial clearly has more 2 2 s than 5 5 s in its factorization so you only need to count how many 5 5 s are there in the factorization of 1000! 1000!
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