Mock AIME 4 2006-2007 Problems/Problem 5

Problem

How many 10-digit positive integers have all digits either 1 or 2, and have two consecutive 1's?

Solution

We take as our universe the set of 10-digit integers whose digits are all either 1 or 2, of which there are $2^{10}$ , and we count the complement. The complement is the set of 10-digit positive integers composed of the digits 1 and 2 with no two consecutive 1s. Counting such numbers is a popular combinatorial problem: we approach it via a recursion.

There are two "good" one-digit numbers (1 and 2) and three good two-digit numbers (12, 21 and 22). Each such $n$ -digit number is formed either by gluing "2" on to the end of a good $(n - 1)$ -digit number or by gluing "21" onto the end of a good $(n - 2)$ -digit number. This is a bijection between the good $n$ -digit numbers and the union of the good $(n-1)$ - and $(n - 2)$ -digit numbers. Thus, the number of good $n$ -digit numbers is the sum of the number of good $(n-1)$ - and $(n - 2)$ -digit numbers. The resulting recursion is exactly that of the Fibonacci numbers with initial values $F_1 = 2$ and $F_2 = 3$ .

Thus, our final answer is $2^{10} - F_{10} = 1024 - 144 = 880$ .

Mock AIME 4 2006-2007 (Problems, Source)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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Mock AIME 4 2006-2007 Problems/Problem 5

Problem

Solution

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