Question by Lê Quốc Trần Anh - Discuss with MathYouLike

Lê Quốc Trần Anh Coordinator

28/12/2017 at 17:57

Given 1989 consecutive numbers from 1 to 1989. Put a "+" or "-" before each number than plus those number we get A. Find the smallest possible positive integer that A can get.

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Lê Anh Duy 28/12/2017 at 22:29

*We have: 1 + 2 - 3 - 4 = - 4

5 + 6 - 7 - 8 = - 4

...

1985 + 1986 - 1987 - 1988 = - 4

=> A = (1 + 2 - 3 - 4) + (5 + 6 - 7 - 8) + ... + (1985 + 1986 - 1987 - 1988) +1989

= - 4 x 497 + 1989 = -1988 + 1989 = 1

* There are 995 odd numbers, 994 even numbers

=> A \(\ne\) 0

=> the smallest possible positive integer that A can get is 1
Lê Quốc Trần Anh selected this answer.
FA Liên Quân Garena 29/12/2017 at 12:28

*We have: 1 + 2 - 3 - 4 = - 4

5 + 6 - 7 - 8 = - 4

...

1985 + 1986 - 1987 - 1988 = - 4

=> A = (1 + 2 - 3 - 4) + (5 + 6 - 7 - 8) + ... + (1985 + 1986 - 1987 - 1988) +1989

= - 4 x 497 + 1989 = -1988 + 1989 = 1

* There are 995 odd numbers, 994 even numbers

=> A ≠≠ 0

=> the smallest possible positive integer that A can get is 1

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