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PDF Editor FAQ

What is the remainder of [math]\sum_{x=0}^{113}(226-2x)!(2x)![/math] when divided by 227?

First things first: [math]227[/math][math][/math] is a prime number, and so, by Wilson’s theorem, [math]226! \equiv -1\pmod{227}[/math].Now,[math](2x)! = 1\cdot 2\cdot 3\ldots (2x) \equiv(-1)^{2x}(227–1)(227–2)(227–3)\ldots(227–2x) = \frac{226!}{(226–2x)!}\pmod{227}[/math].Finally,[math]\sum_{x=0}^{113}(226–2x)!(2x)!\equiv \sum_{x=0}^{113}(226–2x)!\frac{226!}{(226–2x)!} = 113\cdot 226!\equiv114\pmod{227}[/math],and the required remainder is [math]114[/math].

When 3^45 is divided by 4^5, what's the remainder?

To make the numbers more manageable,[math]{3^{45}} \equiv {\left( {{3^9}} \right)^5}\bmod 1024[/math]But [math]3^9=19683[/math] and [math]19683 \equiv 227\bmod 1024[/math].So [math]{3^{45}} \equiv {227^5}\bmod 1024[/math][math]{227^5} \equiv 851\bmod 1024[/math]So the remainder is 851.

What are the ways to express 1000 as the addition of two prime numbers? How do I solve it?

If [math]p[/math] and [math]q[/math] are primes such that [math]p+q = 1000[/math] then one of [math]p[/math] and [math]q[/math] must be less than or equal to [math]500[/math]. Using a simple prime sieve, you find that there are only [math]95[/math] primes less than or equal to [math]500[/math]. For each of these [math]95[/math] primes [math]p_i[/math], test if [math]1000-p_i[/math] is also prime. This gives you the following [math]28[/math] ways to express [math]1000[/math] as the sum of two primes:[math]3+997[/math][math]17+983[/math][math]23+977[/math][math]29+971[/math][math]47+953[/math][math]53+947[/math][math]59+941[/math][math]71+929[/math][math]89+911[/math][math]113+887[/math][math]137+863[/math][math]173+827[/math][math]179+821[/math][math]191+809[/math][math]227+773[/math][math]239+761[/math][math]257+743[/math][math]281+719[/math][math]317+683[/math][math]347+653[/math][math]353+647[/math][math]359+641[/math][math]383+617[/math][math]401+599[/math][math]431+569[/math][math]443+557[/math][math]479+521[/math][math]491+509[/math]If you count [math]3+997[/math] and [math]997+3[/math] as two different ways, then we have [math]56[/math] ways to express [math]1000[/math] as the sum of two primes. Given that there are several solutions, there isn't likely to be an elegant way of finding all of them.

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