AMC 10 · 2011 · #21
Grade 7 arithmeticBrian writes down four integers w>x>y>z whose sum is 44. The pairwise positive differences of these numbers are 1,3,4,5,6, and 9. What is the sum of the possible values for w?
Pick an answer.
AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Brian has four whole numbers in strict decreasing order, $w > x > y > z$, that add up to $44$. If you take the positive difference of every pair of them, the six results are exactly $1, 3, 4, 5, 6,$ and $9$. Add up every value of $w$ that is possible.
Givens: Four integers with $w > x > y > z$; $w + x + y + z = 44$; The six pairwise positive differences are $1, 3, 4, 5, 6, 9$; Answer choices: $16,\ 31,\ 48,\ 62,\ 93$
Unknowns: Every possible value of the largest number $w$, and their sum
Understand
Restated: Brian has four whole numbers in strict decreasing order, $w > x > y > z$, that add up to $44$. If you take the positive difference of every pair of them, the six results are exactly $1, 3, 4, 5, 6,$ and $9$. Add up every value of $w$ that is possible.
Givens: Four integers with $w > x > y > z$; $w + x + y + z = 44$; The six pairwise positive differences are $1, 3, 4, 5, 6, 9$; Answer choices: $16,\ 31,\ 48,\ 62,\ 93$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities
The four numbers only matter through their gaps, so Tool #4 (Introduce a Variable) names the three neighbor gaps $a=w-x$, $b=x-y$, $c=y-z$. Every one of the six pairwise differences is then a run of consecutive gaps, which turns the puzzle into a couple of small equations. Adding all six differences pins down the middle gap $b$ instantly. Tool #2 (Make a Systematic List) then sorts the leftover differences to find the outer gaps, and Tool #3 (Eliminate Possibilities) rules out the arrangements that don't give integers, leaving just the two number sets whose $w$ values we add.
Execute — Answer: B
6.EE.B.6 Step 1 Name the three neighbor gaps
- Instead of tracking the four numbers directly, track the gaps between neighbors: let $a = w-x$, $b = x-y$, and $c = y-z$.
- Every pairwise difference is now a run of these gaps.
- Going from smallest span to largest: $a$, $b$, $c$, then $w-y = a+b$, $x-z = b+c$, and finally $w-z = a+b+c$.
- Since $9$ is the largest listed difference and $w-z$ is the widest span, $w - z = a+b+c = 9$.
💡 The numbers only interact through their gaps, so naming the gaps captures everything that matters.
7.EE.B.4 Step 2 Add up all six differences
- Write the six differences in gap form and add them: $a + b + c + (a+b) + (b+c) + (a+b+c) = 3a + 4b + 3c$.
- That equals $3(a+b+c) + b = 3\cdot 9 + b = 27 + b$.
- But the six given differences add to $1+3+4+5+6+9 = 28$.
- So $27 + b = 28$, which forces the middle gap $b = 1$.
💡 The middle gap $b$ shows up one extra time in the total, so the total minus the rest reveals it.
7.EE.B.4 Step 3 Find the two outer gaps
- With $b=1$ and $a+b+c=9$, the outer gaps satisfy $a + c = 8$.
- The four remaining differences $\{3,4,5,6\}$ must be exactly $a$, $c$, $a+b = a+1$, and $c+b = c+1$.
- Look for two numbers from $\{3,4,5,6\}$ that sum to $8$: only $3$ and $5$ work.
- Check: if $a=3,c=5$ then $a+1=4$ and $c+1=6$, giving $\{3,5,4,6\}$ — exactly the leftover set.
- So $\{a,c\} = \{3,5\}$.
💡 Two gaps and each gap-plus-one must fill four slots, so the pair adding to 8 is the only fit.
6.EE.B.7 Step 4 Turn each gap pattern into numbers
- Two orderings survive.
- Write each number from $w$: $x=w-a$, $y=w-a-b$, $z=w-9$, so the total is $4w-(3a+2b+c)=44$.
- \nFor $a=3,b=1,c=5$: $3a+2b+c = 9+2+5 = 16$, so $4w = 60$ and $w=15$, giving $(15,12,11,6)$.
- \nFor $a=5,b=1,c=3$: $3a+2b+c = 15+2+3 = 20$, so $4w = 64$ and $w=16$, giving $(16,11,10,7)$.
💡 Once the gaps are set, the fixed total $44$ locks the top number to a single value.
7.NS.A.1 Step 5 Add the possible values of w
- Both sets are valid: $(15,12,11,6)$ has differences $\{3,4,9,1,6,5\}$ and $(16,11,10,7)$ has differences $\{5,6,9,1,4,3\}$ — each is exactly $\{1,3,4,5,6,9\}$, and each sums to $44$.
- The possible values of $w$ are $15$ and $16$, so their sum is $15 + 16 = 31$.
- The answer is $\textbf{(B)}$.
💡 Both gap orders produce genuine solutions, so both top values count toward the sum.
6.EE.B.6 Instead of tracking the four numbers directly, track the gaps between neighbors: 7.EE.B.4 Write the six differences in gap form and add them: $a + b + c + (a+b) + (b+c) + 7.EE.B.4 With $b=1$ and $a+b+c=9$, the outer gaps satisfy $a + c = 8$. The four remaining 6.EE.B.7 Two orderings survive. Write each number from $w$: $x=w-a$, $y=w-a-b$, $z=w-9$, 7.NS.A.1 Both sets are valid: $(15,12,11,6)$ has differences $\{3,4,9,1,6,5\}$ and $(16,1 Review
Reasonableness: Both candidate sets pass every test: strictly decreasing integers, sum $44$, and pairwise differences exactly $\{1,3,4,5,6,9\}$. The two tops $15$ and $16$ differ by $1$, which matches $b=1$ being the only gap that changed roles between the two orderings. Their sum $31$ is choice (B); the trap answer $16$ (choice A) is just one of the two $w$ values, and $48$, $62$, $93$ are far too big for a top number that can be at most about $\tfrac{44}{4}+9$.
Alternative: Skip the gap variables and reason on the numbers. From $w-z=9$ and $w+x+y+z=44$, you get $x+y = 44 - (w+z) = 44 - (2z+9) = 35 - 2z$. Testing small integer values of $z$ that keep $w=z+9$ the largest, only $z=6$ and $z=7$ let the in-between values $x,y$ produce the required difference set, again giving $w=15$ and $w=16$ and the sum $31$.
CCSS standards used (min grade 7)
6.EE.B.6Use variables to represent numbers and write expressions when solving a problem (Naming the three neighbor gaps $a,b,c$ and writing each pairwise difference as a run of gaps.)7.EE.B.4Use variables to represent quantities and construct equations to solve problems (Summing the six differences to solve for $b$, and matching the leftover differences to find $a$ and $c$.)6.EE.B.7Solve mathematical problems by writing and solving equations of the form x + p = q and px = q (Using the fixed total $4w-(3a+2b+c)=44$ to solve for $w$ in each ordering.)7.NS.A.1Apply and extend previous understandings of addition and subtraction of integers (Reconstructing each integer set and confirming its pairwise differences and sum.)
⭐ Track the gaps between neighbors instead of the numbers themselves, and every pairwise difference becomes a short sum you can solve.
⭐ Track the gaps between neighbors instead of the numbers themselves, and every pairwise difference becomes a short sum you can solve.
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