Andrew chooses a number from 1 to 100, and Mary also chooses a number from 1 to 100. (They may choose the same number.) It turns out that the product of their numbers is divisible by 3. In how many ways could Andrew and Mary have chosen their numbers?

Guest Nov 16, 2020

#1**+2 **

I will take a crack at this one! Here is my thinking:

Mary and Andrew can choose any number between 1 and 100. The maximum "sample space" of the products of their two numbers should be: 100 x 100 =10,000. So, the question becomes: how many of those 10,000 products are divisible by 3?

I wrote a short computer code to determine that and the answer came out as: 5,511 out of those 10,000 products. So, the conclusion I come to is: there are 5,511 ways that Mary and Andrew could have chosen their numbers.

Note: Melody, CPhill and others should take a good look at my attempt. Is there a simple solution to it, or a simple formula one could use? I don't know!.

Guest Nov 16, 2020

#2**+2 **

Thanks Guest. It is always good to have an answer to compare mine too.

Andrew chooses a number from 1 to 100, and Mary also chooses a number from 1 to 100. (They may choose the same number.) It turns out that the product of their numbers is divisible by 3. In how many ways could Andrew and Mary have chosen their numbers?

Here is the mathematical way.

One or both of them has to choose a multiple of 3. There are 100 numbers and 33 multiples of 3 (for each of them)

There are 33*100 ways for Mary to chose a multiple of 3 and Andrew to chose any number

There are 33*100 ways for Andrew to chose a multiple of 3 and Mary to chose any number

But I have double counted.

There are 33*33 ways that they can both chose a multiple of 3.

**So altogether there are 33*100 + 33*100 - 33*33 = 5511 ways. Just as answering guest found.**

This could be displayed in a ven diagram. 2 overlapping circles.

33*33= 1089 is the overlapping bits

and 100*33 - 1089 = 2211 in each of the other two arias.

Melody Nov 16, 2020