Divisibility by 3 and a Contest

Sums and differences again: remainders

Last time we talked about using sums and differences to tell if a number is divisible by 7.

For example, 781 is not divisible by 7: $781=770+11$, and we know that 770 is divisible by 7 and that 11 is not.

A similar kind of logic works with remainders.

• What is the remainder (R) when you divide 465 by 7?

  $465=420+45$.

  $45÷7$ → R3.

• What is the remainder if you divide (13 + 283) by 7?

• Method 1: $13+283=296=280+16$. $16÷7$ → R2
• Method 2: $13÷7$ → R6; $283÷7$ → R3; $6+3=9$; $9÷7$ → R2, so $(13+283)÷7$ → R2

The divisibility by 3 "trick":

If the digits in a number sum to a multiple of 3, then the number is divisible by 3.

How do we know this works?

First, see if there are any obvious exceptions: does it work for the first few multiples of 3 you can think of? How about the first few non-multiples of 3?

Okay, so you may have persuaded yourself that there is something to it, but does this prove it always works?

Of course not!

Here's the reason it works:

$10÷3$ → R1

So if you have, for example 54, $54=50+4$. You know that 50 will have a remainder of 1 when divided by 3, and that will happen 5 times. As far as remainders are concerned, this is just like adding 5.

We had to stop here to start...

Our first contest

which was really fun!

We'll go over the solutions, not next week, but the week after.