# APMOPS 2013: Đếm số học

Question:

How many different ways are there to select 2 distinct integers  from {2000, 2001, 2002,…,2014, 2015} such that the product of the 2 numbers is divisible by 6? (Note: order is not important, choosing 2001 and 2002 is the same as choosing 2002 and 2001.)

Nhận xét:

Dấu hiệu chia hết cho 6 chúng ta cần nhớ là khi một số chia hết cho cả 2 và 3. Trong bài này ta cần đếm các số chia hết cho 2, các số chia hết cho 3, và các số chia hết cho cả 2 lẫn 3.

Solution:

There are 5 numbers which are divisible by 2 but 3: 2000, 2002, 2006, 2008, 2012, 2014.

There are 3 numbers which are divisible by 3 but 2: 2001, 2007, 2013.

There are  2 numbers which are divisible by both 2 and 3: 2004, 2010.

In order to be divisible by 6, the product have to contain one number which is divisible by 2 and one number which is divisible by 3, or contain one number which is divisible by 6. Hence, there are $2C2+2(16-2)+(8-2)(5-2)=1+28+6\times 3 =47.$ such ways.