2001 AMC 10 Problem 8

P1-10~2 minPrint

GCD & LCM Set up the equation/formula and compute; no special trick needed

Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their schedule is as follows: Darren works every third school day, Wanda works every fourth school day, Beatrice works every sixth school day, and Chi works every seventh school day. Today they are all working in the math lab. In how many school days from today will they next be together tutoring in the lab?

Show solution without answering

Key insight

The four tutors are all together again on the first day that is a multiple of 3, 4, 6 and 7 at once. That is their least common multiple, 84.

Solution

Call today day $0$ . Darren works on days $3, 6, 9, 12, \ldots$ , the multiples of $3$ . In the same way Wanda works on multiples of $4$ , Beatrice on multiples of $6$ , and Chi on multiples of $7$ .
So the next day they are all together is the smallest number that is a multiple of $3$ , $4$ , $6$ and $7$ at the same time. This is called the least common multiple.
Break each number into primes: $3 = 3$ , $4 = 2 \times 2$ , $6 = 2 \times 3$ , $7 = 7$ .
The day must contain every one of those factors. It needs $2 \times 2$ (for $4$ ), a $3$ (for $3$ and $6$ ), and a $7$ (for $7$ ). The $6$ is already covered by the $2$ and the $3$ .
So the day is $2 \times 2 \times 3 \times 7 = 84$ .
Check: $84 \div 3 = 28$ , $84 \div 4 = 21$ , $84 \div 6 = 14$ , $84 \div 7 = 12$ . All whole numbers.

The answer is $\boxed{\textbf{(B)}\ 84}$ .

Why this works

"Every $n$ th day" means the days that are multiples of $n$ , so "everyone together again" means a common multiple, and "next" means the smallest one. The least common multiple is usually smaller than the product, because shared factors are only needed once; here the $6$ adds nothing new.

Alternative approach

Check the choices from smallest to largest. $42$ : $42 \div 4 = 10.5$ , so Wanda is not there. $84$ : divides evenly by $3$ , $4$ , $6$ and $7$ . Since $84$ is the first choice that works, it is the answer.

Concise solution

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

Count days from today. Darren returns on days $3, 6, 9, \dots$ (multiples of $3$ ), Wanda on multiples of $4$ , Beatrice on multiples of $6$ , and Chi on multiples of $7$ . All four are present exactly on the common multiples of $3, 4, 6, 7$ , and the first such day is their least common multiple.

Factor: $3 = 3$ , $4 = 2^2$ , $6 = 2 \cdot 3$ , $7 = 7$ . The lcm takes the largest power of each prime:
$2^2 \cdot 3 \cdot 7 = 84.$

The answer is $\boxed{\textbf{(B)}\ 84}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 3: Multiples and Divisors (§3.3, §3.4, §3.7)
Introduction to Number Theory, Chapter 4: Prime Factorization (§4.7)