AMC 10 · 2010 · #11

Grade 7 probabilitynumber-theory
probability-basicdivisibility-rulesdigit-decomposition casework ↑ Prerequisites: probability-basic
📏 Medium solution 💡 3 insights
Problem
A four-digit palindrome is picked at random. Find the chance it is a multiple of seven.

Pick an answer.

(A)
$\dfrac{1}{10}$
(B)
$\dfrac{1}{9}$
(C)
$\dfrac{1}{7}$
(D)
$\dfrac{1}{6}$
(E)
$\dfrac{1}{5}$
How to solve
Strategy Introduce a Variable

Naming the two free digits turns every palindrome into one algebraic expression. That expression splits into two pieces, and checking each piece for divisibility by 7 is far easier than testing 90 numbers one by one. Counting the winners and dividing by the total then gives the probability.

1STEP 1

Write the palindrome with digits

Two digits build the whole palindrome.

abba = 1000a + 100b + 10b + a = 1001a + 110b
2STEP 2

Split the divisibility into two parts

The outer coefficient already has a factor of seven.

1001 = 7 × 11 × 13 → 7 ∣ 1001a for every a
3STEP 3

Find which inner digits work

So only two inner digits work.

110b ≡ 5b (mod 7), 5b ≡ 0 (mod 7) ⇔ b ∈ {0, 7}
4STEP 4

Count winners over total

The ratio is 1/5, choice (E).

(9 × 2)/(9 × 10) = 18/90 = 1/5
Answer
1/5
Because 1001 already carries the factor of 7, divisibility depends only on the single inner digit b, and 2 out of its 10 values work. A 2-in-10 chance is 1/5, matching choice (E). It also passes a spot check: 7007 (a=7, b=0) = 7 x 1001 and 1771 (a=1, b=7) = 7 x 253 are both palindromes divisible by 7.
💡Key takeaway

Because 1001 is already a multiple of 7, only the middle digit decides the answer, and 2 of its 10 choices work, so the chance is 1/5.

  • Write the palindrome with digits
  • Split the divisibility into two parts
  • Find which inner digits work
  • Count winners over total