62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour

Question

When dividing with remainder some four consecutive positive integers by some three-digit integer it turned out, that the sum of these four remainders is equal to 983. Find the remainder under the division of the smallest of these four numbers by 109.

Accepted Answer

Denote these four consecutive integers by

n

,

n + 1

,

n + 2

and

n + 3

. Denote the three-digit number that we were dividing by as

b

. Let

n = b q + r

. Consider possible values of

r

.

If

r \leq b - 4

, then these remainders are

r

,

r + 1

,

r + 2

and

r + 3 \Rightarrow r + r + 1 + r + 2 + r + 3 = 983 \Rightarrow 4 r = 973

, contradiction.

If

r = b - 3

, then these remainders are

r

,

r + 1

,

r + 2

and

0 \Rightarrow r + r + 1 + r + 2 + 0 = 983 \Rightarrow 3 r = 980

, contradiction.

If

r = b - 2

, then these remainders are

r

,

r + 1

,

0

and

1 \Rightarrow r + r + 1 + 0 + 1 = 983 \Rightarrow 2 r = 981

, contradiction.

If

r = b - 1

, then these remainders are

r

,

0

,

1

and

2 \Rightarrow

r + 0 + 1 + 2 = 983 \Rightarrow r = 980 \Rightarrow b = r + 1 = 981.

It remains to find the required remainder:

n = b q + r = 981 q + 980 = 109 (9 q + 8) + 108.