Question

The set of positive integers is partitioned into $n$ arithmetical progressions with common differences $r_1,r_2,...,r_n$

• A. $N\to \infty$
• B. $N\to 0$
• C. $N\to -\infty$
• D. none of these

Answer 1

The correct option is A $N\to \infty$

Let $a_k=a_1+(k-1)r_1$ he progression with common difference $r_1$. Let us count how many terms of this sequence are less than or equal to some positive integer N. The inequality $a_k\le N$ is equivalent to $a_1+(k-1)r_1\le N$ or,$k\le \frac{N}{r_1}-\frac{a}{r_1}+1.$

It follows that the number of terms of the first progression belonging to the set
$1,2,...N$ equals $|\frac{N}{r_1}-\frac{a}{r_1}+1|.$

Similarly,we deduce that the number of terms of the progression with common difference $r_i$ belonging to the set $1,2,...N|\frac{N}{r_1}-\frac{a}{r_1}+1|.$

Sinced the progessions form a partition of the set of positive integers,we must have ${\sum }_{i=1}^n\lfloor \frac{N}{r_i}-\frac{a}{r_i}+1\rfloor =N.$ Using the inequality $\lfloor x\rfloor \le x\le \lfloor x\rfloor +1,$

we obtain $N\le {\sum }_{i=1}^n(\frac{N}{ri}-\frac{a}{r_i}+1)<N+n$ hence $1\le {\sum }_{i=1}^n\frac{1}{r_i}-\frac{1}{N}{\sum }_{i=1}^n\frac{na}{r_1}+\frac{n}{N}<1+\frac{n}{N}$ and letting $N\to \infty$ yields the desired result.