Number Theory
699.Becausean− 1 ≡n− 1 (modk), the first positive integer greater thanan− 1 that is
congruent tonmodulokmust bean− 1 +1. Thenth positive integer greater thanan− 1 that
is congruent tonmodulokis simply(n− 1 )kmore than the first positive integer greater
thanan− 1 that satisfies this condition. Therefore,an=an− 1 + 1 +(n− 1 )k. Solving this
recurrence gives
an=n+(n− 1 )nk
2.
(Austrian Mathematical Olympiad, 1997)700.First, let us assume that none of the progressions contains consecutive numbers,
for otherwise the property is obvious. Distributing the eight numbers among the three
arithmetic progressions shows that either one of the progressions contains at least four of
the numbers, or two of them contain exactly three of the numbers. In the first situation, if
one progression contains 2, 4 , 6 ,8, then it consists of all positive even numbers, and we
are done. If it contains 1, 3 , 5 ,7, then the other two contain 2, 4 , 6 ,8 and again we have
two possibilities: either a progression contains two consecutive even numbers, whence
it contains all even numbers thereafter, or one progression contains 2,6, the other 4,8,
and hence the latter contains 1980.
Let us assume that two progressions each contain exactly three of the numbers
1 , 2 , 3 , 4 , 5 , 6 , 7 ,8. The numbers 3 and 6 must belong to different progressions, for
otherwise all multiples of 3 occur in one of the progressions and we are done. If 3
belongs to one of the progressions containing exactly three of the numbers, then these
numbers must be 3, 5 ,7. But then the other two progressions contain 2, 4 , 6 ,8, and we
saw before that 1980 occurs in one of them. If 6 belongs to one of the progressions
containing exactly three of the numbers, then these numbers must be 4, 6, 8, and 1980
will then belong to this progression. This completes the proof.
(Austrian–Polish Mathematics Competition, 1980)