AMC 10 · 2002 · #16

Grade 8 arithmetic
Archetype Pair-Sum Invariant └ Sub-archetype Uniform Pair-Sum Recovery
systems-of-equationslinear-equations-two-var convert-to-algebra ↑ Prerequisites: systems-of-equations
📏 Medium solution 💡 2 insights
Problem
For numbers aa, bb, cc, dd, the four expressions a+1a+1, b+2b+2, c+3c+3, d+4d+4 and the sum expression a+b+c+d+5a+b+c+d+5 are all equal to one another. Find the value of a+b+c+da+b+c+d.

Pick an answer.

(A)
-5
(B)
-10/3
(C)
-7/3
(D)
5/3
(E)
5

AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The whole problem hangs on the phrase 'all equal.' Tool #4 (Introduce a Variable) captures that directly: call the shared value k. Every expression in the chain then becomes a short equation involving k, which turns four tangled unknowns into one manageable letter. Tool #15 (Organize Information in More Ways) is the key move: the target a+b+c+d can be written two different ways — once by adding the four individual equations, and once straight from the fifth expression a+b+c+d+5=k. Setting those two versions equal gives a single equation in k. Tool #13 (Convert to Algebra) then finishes it by solving that linear equation.

1STEP 1

Name the shared value

Call the common value kk; each equality becomes its own equation, giving a=k−1a=k-1, b=k−2b=k-2, c=k−3c=k-3, d=k−4d=k-4.

a+1=b+2=c+3=d+4=a+b+c+d+5=k → a=k-1, b=k-2, c=k-3, d=k-4
2STEP 2

Add to get the sum in terms of k

Add the four: the copies of kk stack to 4k4k and the constants give −10-10, so a+b+c+d=4k−10a+b+c+d=4k-10.

a+b+c+d=(k-1)+(k-2)+(k-3)+(k-4)=4k-10
3STEP 3

Read the sum a second way

The chain's last piece says a+b+c+d+5=ka+b+c+d+5=k, so subtracting 55 describes the same sum directly: a+b+c+d=k−5a+b+c+d=k-5.

a+b+c+d+5=k → a+b+c+d=k-5
4STEP 4

Match the two versions and solve for k

One sum cannot be two numbers, so 4k−10=k−54k-10=k-5; tidying up gives 3k=53k=5, hence k=53k=\frac{5}{3}.

4k-10=k-5 → 3k=5 → k=5/3
5STEP 5

Compute the sum

Substitute into a+b+c+d=k−5a+b+c+d=k-5: 53−153\frac{5}{3}-\frac{15}{3} makes the sum −103-\frac{10}{3} — choice (B).

a+b+c+d=k-5=5/3-15/3=-10/3 → (B)
Answer
-10/3
Check every part of the chain with k=5/3: a=5/3-1=2/3, b=-1/3, c=-4/3, d=-7/3. Their sum is (2-1-4-7)/3=-10/3, matching the answer. Testing the chain: a+1=2/3+1=5/3 and a+b+c+d+5=-10/3+5=5/3, so both ends equal k. Everything is consistent, and -10/3 is exactly choice (B).
💡Key takeaway

When a problem says a pile of expressions are all equal, name that shared value with one letter — then write the thing you want two different ways and set them equal.

  • Name the shared value
  • Add to get the sum in terms of k
  • Read the sum a second way
  • Match the two versions and solve for k
  • Compute the sum

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