The set of positive integers is partitioned into n arithmetical progressions with common differences r1,r2,...,rn
A
N→∞
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B
N→0
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C
N→−∞
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D
none of these
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Solution
The correct option is AN→∞ Let ak=a1+(k−1)r1he progression with common difference r1. Let us count how many terms of this sequence are less than or equal to some positive integer N. The inequality ak≤N is equivalent to a1+(k−1)r1≤N or,k≤Nr1−ar1+1.
It follows that the number of terms of the first progression belonging to the set 1,2,...N equals ∣∣Nr1−ar1+1∣∣.
Similarly,we deduce that the number of terms of the progression with common difference ri belonging to the set 1,2,...N∣∣Nr1−ar1+1∣∣.
Sinced the progessions form a partition of the set of positive integers,we must have ∑ni=1⌊Nri−ari+1⌋=N. Using the inequality ⌊x⌋≤x≤⌊x⌋+1,
we obtain N≤∑ni=1(Nri−ari+1)<N+n hence 1≤∑ni=11ri−1N∑ni=1nar1+nN<1+nN and letting N→∞ yields the desired result.