AMC 10 Step by Step

2011 AMC 10B Problem 21

P21-25~6 minPrint

Brian writes down four integers w>x>y>zw > x > y > z whose sum is 4444 . The pairwise positive differences of these numbers are 1,3,4,5,6,1, 3, 4, 5, 6, and 99 . What is the sum of the possible values for ww ?

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Key insight

The biggest difference, 9, is w minus z. It is made of three gaps between neighbors, which must be 1, 3 and 5, with 1 in the middle.

Solution

Picture the four numbers on a number line: zz, then yy, then xx, then ww. Think about the three gaps between neighbors, because every difference is made of them.

1. Name the gaps. Let a=w−xa = w - x, b=x−yb = x - y, c=y−zc = y - z.
2. List all six differences using the gaps.
- Neighbors: aa, bb, cc.
- Two steps apart: w−y=a+bw - y = a + b and x−z=b+cx - z = b + c.
- End to end: w−z=a+b+cw - z = a + b + c.
3. The biggest difference is end to end. So a+b+c=9a + b + c = 9.
4. Which three differences are the gaps? They are three of 1,3,4,5,61, 3, 4, 5, 6 that add to 99.
- If 66 were a gap, the other two would add to 33. No two numbers in the list do that.
- If 44 were a gap, the other two would add to 55. No two of 1,3,5,61, 3, 5, 6 do that.
- That leaves 1+3+5=91 + 3 + 5 = 9. So the gaps are 1,3,51, 3, 5 in some order, and the two-step sums must be the other two numbers, 44 and 66.
5. Which gap is in the middle? The two-step sums are a+ba + b and b+cb + c, which both use the middle gap bb.
- Middle 55: sums 5+1=65 + 1 = 6 and 5+3=85 + 3 = 8. But 88 is not in the list. No.
- Middle 33: sums 3+1=43 + 1 = 4 and 3+5=83 + 5 = 8. No.
- Middle 11: sums 1+3=41 + 3 = 4 and 1+5=61 + 5 = 6. Yes.
So b=1b = 1, and the outside gaps are 33 and 55 in either order.
6. Case 1: gaps 3,1,53, 1, 5 from the top. The numbers are z+9z + 9, z+6z + 6, z+5z + 5, zz. Their sum is 4z+20=444z + 20 = 44, so z=6z = 6. The list is 15,12,11,615, 12, 11, 6, and w=15w = 15.
7. Case 2: gaps 5,1,35, 1, 3 from the top. The numbers are z+9z + 9, z+4z + 4, z+3z + 3, zz. Their sum is 4z+16=444z + 16 = 44, so z=7z = 7. The list is 16,11,10,716, 11, 10, 7, and w=16w = 16.
8. Check Case 1: 15−12=315 - 12 = 3, 12−11=112 - 11 = 1, 11−6=511 - 6 = 5, 15−11=415 - 11 = 4, 12−6=612 - 6 = 6, 15−6=915 - 6 = 9. All six match. Case 2 checks the same way.
9. The sum of the possible values of ww is 15+16=3115 + 16 = 31.

The answer is (B) 31\boxed{\textbf{(B)}\ 31}.

Why this works

For numbers in order, every difference is a sum of neighboring gaps, and the largest difference is the whole spread. Starting from the largest difference shrinks the search to picking three gaps with a fixed total. When a question asks for "the sum of the possible values," expect more than one answer and look for all of them.

Concise solution

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

Write the consecutive gaps as a=w−xa = w - x, b=x−yb = x - y, c=y−zc = y - z. The six pairwise differences are then a,b,c,a+b,b+ca, b, c, a + b, b + c, and a+b+ca + b + c. The biggest of the six is w−z=a+b+cw - z = a + b + c, so a+b+c=9a + b + c = 9.

The three gaps are three of the numbers 1,3,4,5,61, 3, 4, 5, 6 adding to 99. The only such triple is {1,3,5}\{1, 3, 5\} (for instance 4+5=94 + 5 = 9 already, and 3+6=93 + 6 = 9 likewise, leaving no room for a third positive gap).

Now decide the order. The two-step sums a+ba + b and b+cb + c must be the remaining differences 44 and 66. If 55 sat in the middle, one two-step sum would be 5+1=65 + 1 = 6 and the other 5+3=85 + 3 = 8, which is not allowed. So the middle gap b=1b = 1, and (a,c)=(3,5)(a, c) = (3, 5) or (5,3)(5, 3); both give two-step sums 44 and 66.

  • Gaps 3,1,53, 1, 5: the numbers are z+9,z+6,z+5,zz + 9, z + 6, z + 5, z, sum 4z+20=444z + 20 = 44, so z=6z = 6 and w=15w = 15.
  • Gaps 5,1,35, 1, 3: the numbers are z+9,z+4,z+3,zz + 9, z + 4, z + 3, z, sum 4z+16=444z + 16 = 44, so z=7z = 7 and w=16w = 16.

Both lists (15,12,11,615, 12, 11, 6 and 16,11,10,716, 11, 10, 7) really do have differences 1,3,4,5,6,91, 3, 4, 5, 6, 9. The sum of the possible values of ww is 15+16=3115 + 16 = 31.

The answer is (B) 31\boxed{\textbf{(B)}\ 31}.

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.