Difference between revisions of "2025 SSMO Relay Round 1 Problems/Problem 1"
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| + | From the given condition, we know that <math>3</math> divides both <math>x_1+x_2+x_3</math> and <math>x_2+x_3+x_4</math>, so <math>3</math> divides their difference, which is <math>x_4-x_1</math>. Analogous reasoning shows that <math>3</math> divides <math>x_7 - x_4</math> as well. Hence, <math>3</math> divides <math>(x_4 - x_1) + (x_7-x_4) = x_7 - x_1</math>. This means that the smallest possible positive value of <math>x_7 - x_1</math> is <math>3</math>. This minimum is indeed achievable -- take <math>\{x_1, x_2, \dots, x_7\}</math> to be <math>\{0,6,9,12,15,18,3\}</math>, for example -- so the answer is <math>\boxed{3}</math>. | ||
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| + | ~Sedro | ||
Latest revision as of 22:24, 21 October 2025
Problem
Let 
be distinct integers such that the mean of 
is an integer for all integers 
. Find the minimum possible positive value of 
.
Solution
From the given condition, we know that 
divides both 
and 
, so divides their difference, which is 
. Analogous reasoning shows that divides 
as well. Hence, divides 
. This means that the smallest possible positive value of is . This minimum is indeed achievable -- take 
to be 
, for example -- so the answer is 
.
~Sedro
