2010 AMC 10B Problem 8
A ticket to a school play cost dollars, where is a whole number. A group of 9th graders buys tickets costing a total of , and a group of 10th graders buys tickets costing a total of . How many values for are possible?
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Key insight
The ticket price must divide both 48 and 64 evenly. The prices that do are 1, 2, 4, 8 and 16, the numbers that divide 16.
Solution
Each group bought a whole number of tickets at dollars each, so must divide evenly and must also divide evenly.
1. List the numbers that divide :
2. List the numbers that divide :
3. The numbers on both lists are . That is values.
4. Each one really works. For example, at the 9th graders buy tickets and the 10th graders buy ; at they buy and .
The answer is .
Why this works
The numbers that divide both and are exactly the numbers that divide their greatest common factor, . So "divides both" turns into "divides ", and has divisors: . Do not drop the smallest one () or the largest one ().
Concise solution
A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.
Each group spent a whole number of tickets times dollars, so divides and divides . A common divisor of two numbers is a divisor of their greatest common divisor:
The divisors of are , five in all, and each one is a legitimate price (for example means and tickets).
The answer is .
Review in the AoPS books
- Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
- Introduction to Number Theory, Chapter 3: Multiples and Divisors (§3.3, §3.4, §3.7)
- Introduction to Number Theory, Chapter 5: Divisor Problems (§5.2, §5.3)
- Introduction to Number Theory, Chapter 4: Prime Factorization (§4.7)
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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.