AMC 10 · 2004 · #3

Grade 7 algebra
Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search
linear-diophantineconsecutive-integer-count convert-to-algebrabound-inequality-then-enumerate ↑ Prerequisites: linear-equations-two-var
📏 Medium solution 💡 1 insight
Problem
Count the ordered pairs (x,y) of positive integers that satisfy x + 2y = 100.

Pick an answer.

(A)
33
(B)
49
(C)
50
(D)
99
(E)
100
How to solve
Strategy Introduce a Variable

One equation cannot pin down two unknowns, but it can express one in terms of the other. Solving for x turns every choice of y into exactly one value of x (Tool #4), so counting solution pairs becomes the much easier job of counting allowed values of y — one variable instead of two. The two positivity requirements then become a lower and an upper bound on y (Tool #14), and the count is just the number of integers between those bounds (Tool #2). The whole difficulty of the problem lives at the two ends of that range, so both endpoints get checked explicitly. Re-organising the same equation around parity (Tool #15) gives an independent second count that confirms the first.

1STEP 1

Express x in terms of y

Solving for one variable makes the other the only free choice.

x + 2y = 100 ⟹ x = 100 - 2y
2STEP 2

Turn positivity into bounds on y

Both positivity conditions turn into the range 1 to 49.

100 - 2y ≥ 1 ⟹ 2y ≤ 99 ⟹ y ≤ 49.5 ⟹ y ≤ 49
3STEP 3

Check the two ends

Testing the boundaries confirms 49 works but the next value fails.

y=1 → x=98; y=49 → x=2; y=50 → x=0 (rejected)
4STEP 4

Count the values

Counting those whole numbers gives 49 pairs, choice (B).

49 - 1 + 1 = 49 → (B)
Answer
49
A size check: 2y can be at most a little under 100, so y is at most about 50 — meaning the count has to be near 50, not near 100. That immediately makes (D) 99 and (E) 100 implausible, since they would require roughly one solution for every value of x from 1 to 100, and half of those values of x leave 100-x odd, which cannot equal 2y. Between the two survivors, the boundary check settles it: y=49 works, y=50 does not, so the count is 49, one less than the round 50.
💡Key takeaway

With one equation and two unknowns, solve for one and let the other run free — here y can be any integer from 1 to 49, which is 49 pairs, and y=50 fails only because it would make x zero.

  • Express x in terms of y
  • Turn positivity into bounds on y
  • Check the two ends
  • Count the values

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