Download PROBLEM SET 1 1. (i) Prove that for every natural number n 2, one

PROBLEM SET 1
TO BE HANDED IN BY FRIDAY 3TH FEBRUARY 2017
1. (i) Prove that for every natural number n > 2, one has (n + 1) | (n3 + 1).
(ii) Suppose that n is a natural number exceeding 1. Prove that (n2 − 1) |
(n3 + 1) if and only if n = 2.
2. (i) Let a and b be integers. Show that 11 | (100a + b) if and only if
11 | (a + b), and hence deduce that an integer n is divisible by 11 if and only
if the sum of its base-100 digits is divisible by 11.
(ii) Let the conventional base 10 expansion of the integer n be nk nk−1 . . . n1 n0 ,
so that n = 10k nk + 10k−1 nk−1 + . . . + n0 . Let m be the integer with base 10
expansion nk nk−1 . . . n1 , so that m = 10k−1 nk + 10k−2 nk−1 + . . . + n1 . Show
that 2n (and hence also n) is divisible by 19 if and only if m + 2n0 is divisible
by 19, thereby providing a test for divisibility by 19.
3. (i) Let n > 2 be a natural number. Prove
that (n! − 1, (n − 1)! − 1) = 1.
n
n!
(ii) Using that the binomial coefficient
=
is integral, prove
k
k!(n − k)!
that the product of k consecutive integers is always divisible by k!.
4. (i) Apply the Euclidean algorithm to determine (3990, 2016);
(ii) Find integers x and y such that 3990x + 2016y = (3990, 2016).
5. Let n be a natural number and p be a prime number.
(i) Show that for each h ∈ N, the
number of integers m, with 1 6 m 6 n,
n
divisible by ph is equal to
, where bθc denotes the largest integer not
ph
exceeding θ (the integer part of θ).
(ii) Hence deduce that the power of p dividing n! is pep , where ep < n/(p − 1).
6∗ [Hard]. Let 1 < a1 < · · · < ak < 2n be integers not dividing each other.
(i) Show that k 6 n.
(ii) Prove that if k = n and m is the integer satisfying 3m < 2n < 3m+1 then
a1 > 2m .
[Hint: Write each integer ai in the form (2b + 1)2c . In the second part write
a1 = (2m1 + 1)2r and investigate how many numbers ai must be of the form
(2m1 + 1)2c 3d .]
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