2006 AMC 10B Problem 11

P11-15~3 minPrint

Modular Arithmetic Divisibility & Factors Bound the quantity above/below or estimate to pin it down Set up the equation/formula and compute; no special trick needed

What is the tens digit in the sum $7!+8!+9!+...+2006!$

Show solution without answering

Key insight

Every factorial from 10! onward ends in 00, so it cannot change the tens digit. Only 7! + 8! + 9! matters.

Solution

1. The tens digit of a sum only depends on the last two digits of each number. So look for terms that end in $00$ and ignore them.
2. $10! = 1 \cdot 2 \cdot 3 \cdots 9 \cdot 10$ contains the factor $10$ and also the factors $2$ and $5$ , which make another $10$ . So $10!$ is a multiple of $100$ and ends in $00$ .
3. Every later factorial, like $11! = 11 \cdot 10!$ , is a multiple of $10!$ , so it also ends in $00$ . All the terms from $10!$ to $2006!$ add nothing to the last two digits.
4. That leaves only three terms. Work them out:
$7! = 5040,\qquad 8! = 8 \cdot 5040 = 40320,\qquad 9! = 9 \cdot 40320 = 362880.$
5. Only the last two digits matter: $40 + 20 + 80 = 140$ , which ends in $40$ . So the whole sum ends in $40$ .

The tens digit is $4$ .

The answer is $\boxed{\textbf{(C)}\ 4}$ .

Why this works

A question about the last digit or the tens digit only cares about the last one or two digits of each number. Factorials quickly pick up lots of factors of $2$ and $5$ , so from some point on they all end in zeros and drop out. That turns a sum with two thousand terms into a sum of just a few.

Alternative approach

Add the three numbers in full: $5040 + 40320 + 362880 = 408240$ . Its tens digit is $4$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The tens digit depends only on the sum modulo $100$ .

For $n \ge 10$ , the product $n!$ includes the factors $5$ , $10$ (and plenty of $2$ s), so $100 \mid n!$ . Every term from $10!$ onward contributes $0$ modulo $100$ and can be discarded.

That leaves $7! + 8! + 9!$ . Rather than computing the large numbers, factor out $7! = 5040$ :
$7!\,(1 + 8 + 72) = 5040 \cdot 81.$
Modulo $100$ , $5040 \equiv 40$ and $81 \equiv 81$ , so the sum is $\equiv 40 \cdot 81 = 3240 \equiv 40 \pmod{100}$ .

The last two digits are $40$ , so the tens digit is $4$ .

The answer is $\boxed{\textbf{(C)}\ 4}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 12: Introduction to Modular Arithmetic
Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
Introduction to Number Theory, Chapter 10: Units Digits
Introduction to Number Theory, Chapter 5: Divisor Problems (§5.2, §5.3)