AMC 10 · 2025 · #5

Grade 7 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
linear-diophantineparitydivisibility-rules bound-inequality-then-enumeratesystematic-enumeration ↑ Prerequisites: linear-equations-two-var
📏 Medium solution 💡 2 insights
Problem
Two positive whole numbers x and y make the equation 57x + 22y = 400 true. Among every pair that works, find the smallest value the sum x + y can have.

Pick an answer.

(A)
10
(B)
11
(C)
13
(D)
14
(E)
15
How to solve
Strategy Make a Systematic List

The equation alone cannot pin down two unknowns, so the real leverage is the words "positive integers". Because 57 is large compared with 400, only a handful of values of x can possibly fit, so the plan is to trap x inside a short range, throw out the values that cannot work, and then check the few survivors one at a time. Once the complete list of valid pairs is in hand, picking the least sum is just reading the list, and no search for a clever trick is needed.

1STEP 1

Trap x in a short range

With y at least one, x is below seven.

y ≥ 1 ⟹ 57x = 400 - 22y ≤ 378 ⟹ x ≤ 378/57 ＜ 7
2STEP 2

Parity throws out odd x

Parity forces x to be even.

400 even, 22y even ⟹ 57x even; 57 odd ⟹ x even
3STEP 3

Test the three survivors

Of three candidates only two passes.

x = 2: 400 - 114 = 286 = 22 · 13 ✓ x = 4: 172 × x = 6: 58 ×
4STEP 4

Solve for y and verify

Then y is 13 and it checks out.

22y = 400 - 57 · 2 = 286 ⟹ y = 286/22 = 13; 57 · 2 + 22 · 13 = 400
5STEP 5

One pair settles the least

With one pair only, the sum is 15.

x + y = 2 + 13 = 15
Answer
15
The pair checks out directly: 57 · 2 + 22 · 13 = 114 + 286 = 400. The claim that nothing smaller exists survives a wider sweep too. Testing the odd values that parity ruled out, x = 1 leaves 343, x = 3 leaves 229, and x = 5 leaves 115, and none of those is a multiple of 22, so the parity shortcut discarded nothing real. It also explains why the smaller choices look tempting. If y were allowed to be fractional, larger x would give smaller sums: x = 5 gives y = 115/22 and a sum near 10, and x = 3 gives y = 229/22 and a sum near 13. Those near-misses are exactly the values 10 and 13 on the answer list, and every one of them dies on the whole-number requirement.
💡Key takeaway

When one equation holds two unknowns but both must be positive whole numbers, squeeze one variable into a short range first, then check the few numbers left.

  • Trap x in a short range
  • Parity throws out odd x
  • Test the three survivors
  • Solve for y and verify
  • One pair settles the least

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