Olympiad Maths Prep

Track / Stage 4 / 130 of 340 #390 of 2000

Problem 390

AMC 12 late, AIME early
Algebra Difficulty 4.8 Find the answer

10. Given
Sn=n1+2n2++10n10 S_{n}=|n-1|+2|n-2|+\cdots+10|n-10| \text {, }

where, nZ+n \in \mathbf{Z}_{+}. Then the minimum value of SnS_{n} is \qquad

Official solution

10. 112 .

From the problem, we know
Sn+1Sn=n+2n1++10n9[n1+2n2++10n10]=n+n1++n910n10. \begin{array}{l} S_{n+1}-S_{n} \\ =|n|+2|n-1|+\cdots+10|n-9|- \\ {[|n-1|+2|n-2|+\cdots+10|n-10|] } \\ =|n|+|n-1|+\cdots+|n-9|-10|n-10| . \end{array}

When n10n \geqslant 10, Sn+1Sn>0S_{n+1}-S_{n}>0, thus, SnS_{n} is monotonically increasing; \square
When n=0n=0, S1S0>0S_{1}-S_{0}>0;
When n=6n=6, S7S6<0S_{7}-S_{6}<0.
Therefore, SnS_{n} is monotonically decreasing in the interval [1,7][1,7] and monotonically increasing in the interval [7,+)[7,+\infty).
Hence, the minimum value of SnS_{n} is S7=112S_{7}=112.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.