By Andrew Granville
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This booklet is an elevated model of lectures given at a summer time university on symplectic geometry in Nordfjordeid, Norway, in June 2001. The unifying function of the publication is an emphasis on Calabi-Yau manifolds. the 1st half discusses holonomy teams and calibrated submanifolds, concentrating on precise Lagrangian submanifolds and the SYZ conjecture.
6-page laminated consultant comprises: ·general phrases ·lines ·line segments ·rays ·angles ·transversal line angles ·polygons ·circles ·theorems & relationships ·postulates ·geometric formulation
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1 · 2 · 3 · · · (n − 1) · n. If pk is the power of p dividing m then we will count 1 for p dividing m, then 1 for p2 dividing m, . . , and finally 1 for pk dividing m. Therefore the power of p dividing n! equals the number of integers m, 1 ≤ m ≤ n that are divisible by p, plus the number of integers m, 1 ≤ m ≤ n that are divisible by p2 , plus etc. The result then follows from the previous exercise. Kummer’s Theorem. The largest power of prime p that divides the binomial coefficient a+b is given by the number of carries when adding a and b in base p.
Mk are a set of pairwise coprime positive integers. For any set of residue classes a1 (mod m1 ), a2 (mod m2 ), . . , ak (mod mk ), there exists a unique residue class x (mod m), where m = m1 m2 . . mk , such that x ≡ aj (mod mj ) for each j. Proof. We can map x (mod m) to the vector (x (mod m1 ), x (mod m2 ), . . , x (mod mk )). There are m1 m2 . . 17(a). Hence there is a suitable 1-to-1 correspondence between residue classes mod m and vectors, which implies the result. This is known as the Chinese Remainder Theorem because of the ancient Chinese practice (as discussed in Sun Tzu’s 4th century Classic Calculations) of counting the number of soldiers in a platoon by having them line up in three columns and seeing how many are left over, then in five columns and seeing how many are left over, and finally in seven columns and seeing how many are left over, etc.
Divisors using factorizations. Suppose that n = bp 23 If n = ab then pp . p prime pnp , a = p pap , b = 2n2 3n3 5n5 · · · = 2a2 3a3 5a5 · · · 2b2 3b3 5b5 · · · = 2a2 +b2 3a3 +b3 5a5 +b5 · · · , 23 Here, and often hereafter, we suppress writing “prime” in the subscript of , for convenience. 28 ANDREW GRANVILLE so by the fundamental theorem of arithmetic we have np = ap + bp for each prime p. As ap , bp ≥ 0 for each prime p, we can deduce that 0 ≤ ap ≤ np . In the other direction if a = 2a2 3a3 5a5 .