What is $1000 choose 326$ $mod {13}$?
The exponent of 13 on the factorisation of $1000!$ is $lfloorfrac {1000} {13}rfloor+lfloorfrac {1000} {13^2}rfloor$ do the same for $326!$ and $674!$ and you''ll find that
statistics
Imagine I asked 1000 1000 people to choose a number between 0 0 and 999 999 (both inclusive, the numbers are not biased, they will be completely random) and write that number down.
$1000$ small cubes are assembled into a larger cube. If one layer of
1000 1000 is the number of small cubes in the original cube. Each face of the original cube contains 10 × 10 = 100 10 × 10 = 100 small cubes, so the effect of removing the small cubes on all six
algebra precalculus
For example, the sum of all numbers less than 1000 1000 is about 500, 000 500, 000. So, 168 1000 × 500, 000 168 1000 × 500, 000 or 84, 000 84, 000 should be in the right ballpark. 76127
Find the remainder when $7^{7^{7}}$ is divided by 1000
Find the remainder when 777 7 7 7 is divided by 1000 Ask Question Asked 8 years, 6 months ago Modified 8 years, 6 months ago
combinatorics
The number of bacteria in a culture is 1000 and this number increases by 250% every two hours. How many bacteria is present after 24 hours?
combinatorics
Number of ways to invest $20, 000 $ 20, 000 in units of $1000 $ 1000 if not all the money need be spent Ask Question Asked 2 years, 11 months ago Modified 2 years, 11 months ago
Why is kg/m³ to g/cm³1 to 1000?
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
arithmetic
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
Creating arithmetic expression equal to 1000 using exactly eight 8''s
I would like to find all the expressions that can be created using nothing but arithmetic operators, exactly eight $8$''s, and parentheses. Here are the seven solutions I''ve found (on the Internet)...
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