Concrete mathematics : a foundation for computer science

A ANSWERS TO EXERCISES 507
this is divisible by 11.
Therefore the answer is: Either n = 1 or n # 4 is composite.
4.54 EJ (1 OOO!) > 500 and es (1 OOO!) == 249, hence 1 OOO! = a. 1 0249 for some
even integer a. Since 1000 = (1300)5, exercise 40 tells us that a. 2249 =
looo!/5249 E -1 (mod 5). Also 2249 = _ 2, hence a = 2, hence a mod 10 = 2
or 7; hence the answer is 2.1 0249.
4.55 One way is to prove by induction that P&Pt(n + 1) is an integer;
this stronger result helps the induction go through. Another way is based
on showing that each prime p divides the numerator at least as often as it
divides the denominator. This reduces to proving the inequality
k=l
which follows from
k=l
[(In - 1 )/ml + LWmj 3 lnhl
The latter is true when 0 6 n < m, and both sides increase by 4 when n is
increased by m.
4.56 Let f(m) = ~~~~’ min(k,2n-k)[m\k], g(m) = EL=: (2n-2k-1) x
[m\(2k+ l)]. The number of times p divides the numerator of the stated
product is f(p) + f(p2) + f(p3) + ..., and the number of times p divides the
denominator is g(p) + g(p2) + g(p3) + ... . But f(m) = g(m) whenever m
is odd, by exercise 2.32. The stated product therefore reduces to 2”‘” ‘1, by
exercise 3.22.
4.57 The hint suggests a standard interchange of summation, since
x [d\ml = x [m= dkl = Ln/dj .
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