AMC 10 · 2013 · #22

Grade 7 probability

Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count

place-valuemulti-digit-arithmeticdigit-constraintsprobability-basic convert-to-algebrawork-backwardssystematic-enumeration ↑ Prerequisites: place-valueprobability-basic

📏 Long solution 💡 4 insights

Problem

A random six-digit palindrome is divided by eleven. Find the chance the result is a palindrome too.

Pick an answer.

(A)

$\frac{8}{25}$

(B)

$\frac{33}{100}$

(C)

$\frac{7}{20}$

(D)

$\frac{9}{25}$

(E)

$\frac{11}{30}$

How to solve

Strategy Introduce a Variable

Tool #4 (Introduce a Variable) is primary: naming the digits turns a statement about two written numbers into arithmetic on letters, which is the only way to control both the numerator and the quotient at once. Tool #11 (Work Backwards) supplies the key reversal — dividing a 6-digit number by 11 is awkward, but multiplying a candidate quotient by 11 is just 10m + m, one addition with visible columns, so we study the map upward instead of downward. Tool #14 (Extreme Principle) pins the quotient between its smallest and largest possible values, which tells us how many digits it can have and lets us close off the case the size bound almost hides. Tool #2 (Make a Systematic List) then counts the surviving digit triples in an organized sweep instead of one by one.

1STEP 1

Name the digits and size the sample space

Three digits describe the whole palindrome.

n = X · 100001 + Y · 10010 + Z · 1100 = 11(9091X + 910Y + 100Z)

2STEP 2

Bound the quotient's size

The quotient's size is quickly bounded.

9091 ≤ m ≤ 90909

3STEP 3

Rule out a 4-digit quotient

A four-digit quotient is impossible.

m = 9QQ9 → n ≡ 9 (mod 10), 100001 ≤ n ≤ 109989 → n starts with 1

4STEP 4

Multiply the candidate quotient by 11

Multiplying back shows the digit sums.

11m = A · 10⁵ + s · 10⁴ + t · 10³ + t · 10² + s · 10 + A, s = A+B, t = B+C

5STEP 5

No carries means it works

With no carry the mirror survives.

A+B ≤ 9 and B+C ≤ 9 → n = A (A+B) (B+C) (B+C) (A+B) A

6STEP 6

A carry always breaks the mirror

A carry always breaks it.

n/11 is a palindrome ⇔ A+B ≤ 9 and B+C ≤ 9

7STEP 7

Count the digit triples

Counting the good triples gives 330.

Σ_B=0⁸ (9-B)(10-B) = 90 + 72 + 56 + 42 + 30 + 20 + 12 + 6 + 2 = 330

8STEP 8

Divide favorable by total

Dividing gives 11/30, choice (E).

P = 330/900 = 11/30 → (E)

Answer

11/30

The value 11/30 ≈ 0.367 sits just above 1/3, which fits: the two conditions A+B ≤ 9 and B+C ≤ 9 each pass comfortably more than half the time, and they overlap heavily through the shared digit B. The count also survives a second summation: grouping instead on d = B and using Σ_B=0⁸(9-B)(10-B) = 2[C(2, 2) + C(3, 2) + … + C(10, 2)] = 2C(11, 3) = 330 by the hockey-stick identity. Spot checks confirm both directions: 123321 ÷ 11 = 11211, a palindrome, and its triple (A,B,C) = (1,1,2) satisfies A+B = 2 ≤ 9, B+C = 3 ≤ 9; while 981189 ÷ 11 = 89199, not a palindrome, matching the fact that no legal triple produces it. Finally 330/900 must reduce to a fraction whose denominator divides 900, and 30 does.

💡Key takeaway

Multiplying a palindrome by 11 just adds it to a shifted copy of itself, so the mirror pattern survives exactly when no column adds past 9 — carries are the only thing that can break it.

Name the digits and size the sample space
Bound the quotient's size
Rule out a 4-digit quotient
Multiply the candidate quotient by 11
No carries means it works
A carry always breaks the mirror
Count the digit triples
Divide favorable by total