Singapore Mental Math (7) – Finding remainders when dividing by …

[Part of the mental math series…]  Here are the strategies for Week 17, 18 & 19 in Grade 6.

Good things:

• I usually see these as divisibility rules, and while it is obvious that the divisibility rules will give the remainder (perhaps with some tweaking), this isn’t commonly explored (in my classroom at any rate), so I found these interesting.
• There is some mental maths to do.
• There is proper use of vocabulary.

Bad things:

• These are incredibly restricted and over-specify the conditions. 
• There is a missed opportunity to look at why they work.
• Some of them are expressed very badly.

Let’s look at each in turn:

Dividing by 5

I hope many pupils would spot that there is a far easier way to deal with the Note!  Rather than dividing the last two digits by 5, why not just instruct pupils to look at the final digit.  In fact, pupils might come up with an easier way to do this.  Let’s ask them to explore how we can tell what the remainder is when we divide by 5.  They could try a few out for themselves, could have an idea and might then link it to the rule for divisibility by 5.

If an integers ends in 0 or 5 then it is divisible by 5.  If it ends in 1 or 6 then the remainder is 1, if it ends in 2 then …

Dividing by 8

This only helps if you have a number with 4 digits or more. 

Why does it work?  1000 is divisible by 8, so any 1000s will automatically be divisible by 8.  Given that we only want the remainder we could also get rid of anything that is clearly a multiple of 8.  To find the remainder when 4169 is divided by 8 we can ignore the thousands, leaving 169.  160 is divisible by 8, so we can ignore that, leaving 9.  The remainder is therefore 1.

(It is a little frustrating that in all of the questions on these three strategies none of the numbers are actually divisible by the divisor.  It would be nice to have a remainder of 0 once in a while.)

Dividing by 9

Many pupils will recognise the algorithm for determining whether a number is a multiple of 9. 

This is phrased rather oddly in that it is only used here for 4-digit numbers (all of the questions are of that form), when it works for any integer.  It also doesn’t explain what happens if the single digit is 9.

Can pupils explain why this algorithm doesn’t produce zero (unless you start with the number 0)?

Can they explain why this algorithm works?  Here is an algebraic way to show it.

If a number looks like “abcd” then the algebra is actually 1000a + 100b + 10c + d.

This can be rewritten as 999a + a + 99b + b + 9c + c + d

Rearrange this to give (999a + 99b + 9c) + (a + b + c + d).

The first bracket is clearly divisible by 9, so we only need to worry about the second one, which is the sum of the digits.