AMC 10 Step by Step

2014 AMC 10A Problem 10

P1-10~1 minPrint

Five positive consecutive integers starting with aa have average bb . What is the average of 55 consecutive integers that start with bb ?

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Key insight

The average of five consecutive whole numbers is the middle one. So b is a + 2, and the second list averages b + 2, which is a + 4.

Solution

Try a small case first. If a=1a = 1, the five numbers are 1,2,3,4,51, 2, 3, 4, 5. Their sum is 1515 and their average is 33, which is the middle number. That always happens: the numbers are spread evenly around the middle one, so the ones below it and above it balance out.

  1. The first list is a,a+1,a+2,a+3,a+4a, a+1, a+2, a+3, a+4. Its middle number is a+2a + 2, so b=a+2b = a + 2.
  2. The second list starts at bb: b,b+1,b+2,b+3,b+4b, b+1, b+2, b+3, b+4. Its average is its middle number, b+2b + 2.
  3. Put in b=a+2b = a + 2: the average is (a+2)+2=a+4(a + 2) + 2 = a + 4.

Check with a=1a = 1: then b=3b = 3, and the list 3,4,5,6,73, 4, 5, 6, 7 has average 55, which is 1+41 + 4.

The answer is (B) a+4\boxed{\textbf{(B)}\ a+4}.

Why this works

When numbers go up by the same step and there is an odd number of them, the average is the middle one, because each number below the middle is matched by one the same distance above it. Using that fact twice makes the problem a two-line calculation. Checking with a small value like a=1a = 1 is a quick way to be sure.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The five integers a,a+1,a+2,a+3,a+4a, a+1, a+2, a+3, a+4 are evenly spaced, so their average is the middle one: b=a+2b = a + 2.

The five consecutive integers starting with bb are b,b+1,b+2,b+3,b+4b, b+1, b+2, b+3, b+4, whose average is likewise the middle term b+2b + 2. Substituting b=a+2b = a + 2,
b+2=(a+2)+2=a+4.b + 2 = (a + 2) + 2 = a + 4.

The answer is (B) a+4\boxed{\textbf{(B)}\ a+4}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.