AMC 10 · 2002 · #15

Grade 7 number-theory
Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Integrality / Divisibility Screen
place-valuedigit-decompositionlinear-equations-two-var convert-to-algebrabound-inequality-then-enumerate ↑ Prerequisites: place-valuedigit-decomposition
📏 Medium solution 💡 2 insights
Problem
Take a four-digit number and cross off its leading digit, leaving a three-digit number. Count the four-digit numbers for which the leftover three-digit number is exactly one ninth of the original.

Pick an answer.

(A)
4
(B)
5
(C)
6
(D)
7
(E)
8
How to solve
Strategy Convert to Algebra

Chopping off a digit sounds like an operation on symbols, but in base ten it is plain subtraction: erasing the leading digit a of a four-digit number removes exactly 1000a. Tool #13 (Convert to Algebra) turns that sentence into an equation. Tool #4 (Introduce a Variable) keeps the bookkeeping light by using only two unknowns — the leading digit and the whole three-digit tail — instead of four separate digits. Tool #14 (Extreme Principle) then finds where the leading digit runs out of room, since the tail is forced to be 125a and must stay under 1000. Tool #2 (Make a Systematic List) writes out the surviving numbers so the count is checked, not assumed.

1STEP 1

Turn deleting a digit into subtraction

With leading digit a and tail x, place value gives N = 1000a + x and the rule says N = 9x.

N = 1000a + x, N = 9x
2STEP 2

Solve for the tail

Equating and simplifying gives x = 125a, so the tail is fully determined by the first digit.

1000a + x = 9x → 1000a = 8x → x = 125a
3STEP 3

See where the digit runs out

The tail must stay under 1000, so 125a at most 999 forces a at most 7.

100 ≤ 125a ≤ 999 → 1 ≤ a ≤ 7
4STEP 4

List them and confirm the count

Each of a = 1 to 7 gives a valid N = 1125a, so the count is 7, choice (D).

N = 1125a: 1125, 2250, 3375, 4500, 5625, 6750, 7875 → 7 → (D)
Answer
7
The relation N = 9x already limits things: x runs over three-digit numbers, so N = 9x runs from 900 to 8991, and N must also be four-digit. That alone would allow many N; the extra force comes from N's tail having to be x itself, which pins N to the multiples of 1125. The multiples of 1125 below 10000 are 1125, …, 7875 and then 9000, and 9000 fails because deleting its leading 9 leaves 000, not a three-digit number. Seven survive, so (D) is right and (E) 8 is exactly the trap of counting 9000.
💡Key takeaway

Deleting the front digit just subtracts its place value, so "one ninth of N" becomes an equation that forces the rest of the number.

  • Turn deleting a digit into subtraction
  • Solve for the tail
  • See where the digit runs out
  • List them and confirm the count

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