Number theory

<1> Let a,b,m,n belongs to Z with m>0 and n|m. show that if


a = b (mod m), then a = (b mod n).

<2> Let a,b,m,n belongs to Z with m>0, n>0 and gcd (m,n)
=1.show that if a = b (mod m),a = b (mod n)then a= b
(mod mn).

<3> Prove that for any prime p and integer x, if x^2 = 1 (mod p)then x = (1 mod p)or x = -1 (mod p).

<4> Show that there are 14 distnict, possible,yearly (gregorian)calenders, show that all 14 calendears actually occur.

=========== thank you so much for spending your time for it============
prameela"

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answer to 1: a = b (mod m) ==> a = mx+b (x belongs to Z)

 , n|m ==> m= dn (d belongs to Z) ==> a= (dn)x+b= (dx)n+b (dx belongs to Z) ==>

a = (b mod n) 

 

 


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<2> if we take for granted that every number can be decomposed as product of primes in an unique way - except the order:

m divides (a-b) hence a-b = m*q

n divides (a-b) hence all the prime factors in n are either in m or in q.

gcd(m,n)=1, so no factor of n is in m. Then all factors of n are in q, or q=n*r for some other integer r. Replacing in the first equality: a-b = m*n*r, proving a = b (mod m*n)

  

<3> x^2-1 = (0 mod p), or (x-1)*(x+1) = pq for some integer q. Since p is prime it must appear in at least one of the decompositions: of (x-1) or of (x+1). Hence p divides (x-1) or (x+1)

 

<4>

For each 365 day year there are 7 possible calendars, depending on which of the 7 days of the week falls on January 1st. For leap years - 366 days - the same argument applies, yielding 7 more combinations. These later combination are different from any of the first 7 because if January 1st happens to be the same, March 1st is necessarily one day later. So there are 14 distinct calendars.

 

 To check that all combinatios will be - were already - realised, take periods of 4 years each starting January 1st. In any such period there are always 1461 days, which decomposes as 7*209 - 2. Hence if the January 1st of a period is Monday - for example - 4 years later January 1st will be Saturday, exactly two week days behind. Cycling through more 4 year periods we can find years that start with any given day of the week; this is known because 7 is a prime number, but in this case it is easy to convince oneself that it is true, through a method similar to counting on one's fingers:

Monday>Saturday>Thursday>Tuesday>Sunday>Friday>Wednesday 

 We also notice that we only need 7 fingers, so 7 periods of 4 years are enough to find all possible 7 calendars for non-leap years. For leap years it suffices to shift the eriods in consideration so that they start with a leap year.

 

Finally, the Gregorian calendar makes exception to the rule that every fourth year is a leap year, but at most once 100 years. As shown above, we can find all 14 calendar kinds within any 31 years period inside a 100 years where there are no exceptions.


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In this delicious implementation of the present god-eat-god multiverse, we offer world-class opinions and expertise for you on a lot of stuff, just ask.  if we don't know, then you forget about it...

too easy.  You can also conveniently prove it's negation as false.  ask a more challenging one and we may reply.


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