Prove that among any 39 consecutive positive integers there exists at least

1.4.8. Prove that among any 39 consecutive positive integers there exists at least

one having the sum of its digits a multiple of 11 .

Solution. Let s n denote the sum of digits of the positive integer n. If the last digit of n is not 9, then s n +1 =s n + 1. If the last k digits of n are 9, then

36 1 Sequences s n +1 =s n − 9k + 1. Next, we take into account that in a sequence of fewer than

100 consecutive positive integers there exists at most one number that ends with at least two digits 9. The most unfavorable case for our problem is the following: s n ,s n + 1, . . . , s n + 8, s n ,s n + 1, . . . , s n + 9, s n − 9k + 10,s n − 9k + 11,...,s n − 9k +

19 ,s n − 9k + 11,s n − 9k + 12,...,s n − 9k + 20, where n − 1 ends with exactly one 0, n + 18 has the last k digits equal to 9, and

−9k ≡ 1 (mod 11) and s n −1 ≡ 0 (mod 11).

In the above scheme, which contains the sum of digits of 39 consecutive positive integers, we observe that only the number s n +38 =s n − 9k + 20 is a multiple of 11. ⊓ ⊔

The existence of the maximum in the next example follows by standard compact- ness arguments. However, it is much more difficult to find the explicit value of the maximum.

1.4.9. Let n ≥2

be an integer, and a 1 ,...,a n real numbers belonging to [−1,1] . Establish for what values of a 1 ,...,a n the expression

a 1 + ··· + a n = −1 1 − + ··· + n

E a 2 + ··· + a n

n achieves its maximum.

Solution. By virtue of the symmetry of E, we can assume that a 1 ≥a 2 ≥ ··· ≥ a n . Define s k =∑ k i =1 a i (k = 1, . . . , n). It follows that

|(n + 1)a −s |.

i =1 n

Let α

be the number of terms a i such that (n + 1)a i ≥s n . Hence

E = [(n + 1)s α − α s n − (n + 1)(s n −s α ) + (n − α )s n ]

1 = [2(n + 1)s n

α −s n (2 α + 1)] .

Assume α is fixed. Since −1 ≤ a i ≤ 1, it follows that s n ≥s α + α − n, s α ≤ α . Therefore

α − (s α + α − n)(2 α + 1)], and thus

E ≤ [(2n + 2)s n

E ≤ [(2n − 2 α + 1) α −( α − n)(2 α + 1)] = n

n (−4 α 2 + 4n α + n). This means that for any fixed α , we have max E = (−4 α 2 + 4n α + n)/n, which is

achieved for a i = 1 (1 ≤ i ≤ α ) and a j = −1 ( j = α + 1, . . . , n). If α is varying, then by monotony arguments, we obtain that the maximum of E is attained for

1.4 Qualitative Results 37

α = n/2 (if n is even) and α = (n + 1)/2 or α = (n − 1)/2 (if n is odd), that is, for α = [n + 1/2]. Therefore

2 2 +1.⊓ ⊔ The solution of the next problem relies deeply on the order structure of real num-

max E =−

bers combined with a certain basic property of the set of positive integers. What is this property and why does the result not remain true if the three sequences are in R?

1.4.10. Prove that for any sequences of positive integers (a n ) n ≥0 , (b n ) n ≥0 , (c n ) n ≥0 there exist different positive integers p and q such that a p ≥a q , b p ≥b q , and c p ≥c q .

Solution. Let q ∈ N be such that c q = min n c n . If there exists p that a p ≥a q and b p ≥b q , then we conclude the proof. Let us now assume that for all p

p <a q or b p <b q . This means that there are infinitely many indices p

p <a q or b p <b q . We can assume, without loss of generality, that a p <a q , for infinitely many p ∈ N. Since between 1 and a q there are finitely many integers, it follows (by Dirichlet’s principle) that between 1 and

a q there exists a number that appears infinitely many times in the sequence (a n ) n ≥0 . In other words, the sequence (a n ) n ≥0 contains a constant subsequence (a n k ): a n 1 =

a n 2 = ··· = a n k = ···. Next, we consider the corresponding subsequences (b n k ) and (c n k ). Let j ∈ N

be such that c n j = min k c n k . If there is some i ∈ N such that b n i ≥b n j , then the proof is concluded because a n i =a n j . So, we assume that b n i <b n j , for all i ∈ N. In particular, the subsequence (b n k ) is bounded, that is, it contains a constant subse- quence (b m k ).

Consider now the subsequence (c m k ). The proof is concluded after observing that choosing r , s ∈ N such that c m r ≥c m s , we have b m r =b m s and a m r =a m s . ⊓ ⊔

Kepler’s equation is an important tool for understanding major phenomena aris- ing in celestial mechanics. This contribution is due to the German mathematician Johannes Kepler (1571–1630), who formulated the three laws of planetary motion,

a major step in the formulation of the laws of motion and universal gravitation by Sir Isaac Newton. The unknown of the following equation denotes the eccentric anomaly , which appears in astrodynamics and is related to Kepler’s second law:

A line joining a planet and the sun sweeps out equal areas during equal intervals of time .