2010 AMC 12B Problem 14

Attempt Problem 14 of the 2010 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 12B solutions, or check the answer key.

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14.

Let

a,

b,

c,

d,

and

e

be positive integers with

a+b+c+d+e=2010,

and let

M

be the largest of the sums

a+b,

b+c,

c+d,

and

d+e.

What is the smallest possible value of

M?

670

671

802

803

804

(a+b)+c+(d+e)=2010,

and each of

a+b,

c,

and

d+e

is at most

M

This forces

3M\ge2010;

rule out

M=670,

then build an example reaching the bound

Each of

a+b,

d+e,

and

c

is at most

M

(note

c\le c+d\le M

). Adding,

2010=(a+b)+c+(d+e)

\le3M,

M\ge670.

M=670,

then

c=670,

but then

b+c\ge671\gt M,

a contradiction. Hence

M\ge671.

The value

671

is reached by

(a,b,c,d,e)=

(669,1,670,1,669),

whose consecutive-pair sums are

670,671,671,670.

Thus, the correct answer is B.

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