Been working on a problem with my 11 year old. It was to find the probability that a three digit number ending in 6, will be a multiple of 6?
Intuitively what do you think this will be? Before reading on? 1/2, 1/4 ?
Anyway, the way we looked at it was to think about the first 10 or so 3 digit numbers which end in 6 - 106, 116, 126 etc and then look at the sum of the digits to see if they were a multiple of 6. We extended it into a grid, like so...
We discovered there were 30 out of 90. So the probability was 1/3.
It felt a little unsatisfactory as we essentially had to search through a list, yes there is a pattern and it wasn't difficult to count but we basically counted them. Is there a quicker way this could be done? Feels like, given the answer comes to 1/3?
Looking beyond an 11 year old, i thought about using some algebra, so a 3 digit numbers abc could be written as 100a+10b+6 and this will be a multiple of 6 if a+b is a multiple of 3 . So i end up looking at solving a+b=3k, where a,b, k are natural numbers.
Suppose k=1: then a=1, b=2 or a=2, b=1
Suppose k=2: then a=1, b=5 or a=5, b=1 AND a=2, b=4 etc...
Again feels a little cumbersome?
Thoughts welcome.
My son then asked me how the probability varied for other last digits. I was confidently able to tell him that if it ended in a 2 it was a multiple of 2 (!).
Any thoughts?
Intuitively what do you think this will be? Before reading on? 1/2, 1/4 ?
Anyway, the way we looked at it was to think about the first 10 or so 3 digit numbers which end in 6 - 106, 116, 126 etc and then look at the sum of the digits to see if they were a multiple of 6. We extended it into a grid, like so...
We discovered there were 30 out of 90. So the probability was 1/3.
It felt a little unsatisfactory as we essentially had to search through a list, yes there is a pattern and it wasn't difficult to count but we basically counted them. Is there a quicker way this could be done? Feels like, given the answer comes to 1/3?
Looking beyond an 11 year old, i thought about using some algebra, so a 3 digit numbers abc could be written as 100a+10b+6 and this will be a multiple of 6 if a+b is a multiple of 3 . So i end up looking at solving a+b=3k, where a,b, k are natural numbers.
Suppose k=1: then a=1, b=2 or a=2, b=1
Suppose k=2: then a=1, b=5 or a=5, b=1 AND a=2, b=4 etc...
Again feels a little cumbersome?
Thoughts welcome.
My son then asked me how the probability varied for other last digits. I was confidently able to tell him that if it ended in a 2 it was a multiple of 2 (!).
Any thoughts?
