Hi folks. My year 5 (age 9) son has a maths problem: Subtract a 3 digit number from a four digit number to give a 3 digit answer. Use each digit 0-9 once only in your solution.

I’m stumped on a method that doesn’t just use trial and error. Apart from 1St digit of 4 digit number being 1, how do we go about solving this? Thanks all!

Oops!

Double oops!

I’m stumped on a method that doesn’t just use trial and error.

What’s wrong with trial and error? I’m pretty sure that was part of maths when I was a kid.

Aged 9? Have they been doing coursework in school to show them how to solve this?

I’d have thought trial an error would be the only sensible approach at that age. Cut out cards with each digit written on them, draw squares on a sheet of A4 to placehold the cards in the sum, then spend the evening shuffling them to find an answer that works (I expect there will be multiple solutions).

ignote that didn’t read it right! Can only see trial and error!

They want trial and error to make them do lots of sums.

1023-456
1024-356
1025-346

…This could be a long evening. 😆

it’s not looking for 1 correct answer, it’s a way of practising column subtraction (show working no calc)

trial and error do lots( well 10)

Mother in law was a principal maths teacher … She can’t think of a simple method without there being pages and pages of algebra.

This isn’t the same sum, but looks to be a similar concept.

http://mathematicscentre.com/taskcentre/043numbr.htm

1089 – 764 = 325

OP is asking for a method, not a solution.

The kid can’t go back to school tomorrow with a correct answer and when asked how he did it say “my dad asked a bunch of alleged cyclists for an answer.”

They want trial and error to make them do lots of sums.

Yes and no. It’s a logic problem more than anything. After a bit of trial and error, you’ll realise that carrying is allowed (the problem doesn’t say it’s not!) and then you can start working out where the zero can go.

At higher level this is done algebraically and you work out that the digits of the 4-digit number always add up to either 9 or 18.
A further concept that builds on this is that 0+1+2+3….+9 = 45 (again, you can do that with algebra)

1062
– 479
583

1206
– 347
859

1305
– 879
426

(there are several other solutions).

Think about where the zero can go, that’s how you start this.

It can’t be placed in the ones or tens places of the number being subtracted. If it was then [x] minus zero = [x] and we can’t have 2 [x]
It can’t be placed in the hundreds place of the number being subtracted either, because then it would be a two digit number instead of a three digit number.
It can’t be in the hundreds place of the answer line for the same reason as ^^.
Also, to get a 0 in the tens or ones place of the answer the two digits above have to be the same, and we don’t have two digits the same to use, so there can’t be a 0 anywhere in the answer line.

Go!

Here’s another

1089
– 456
732

I think logic is supposed to be the orderof the day.

Change ine at a time.

1234-567=
1234-675=
1234-765=

Some of them are become clearer when you start.

It has to be 1000 and something.
You can never end on a 4 and subtract anything that ends in a seven. 8-9 6-8 4-7 2-6 similarly.

You need to start with 10 as its the only number that you can recreate by subtraction maybe?

Are others reading the question the same way as me, in that the answer to the subtraction can include digits used once for the four digit and three digit numbers used in the subtraction?

Or if I use the digits 0-6 inclusive for the initial numbers, the answer can only use the digits 7-9 inclusive?

No, the answer must use unique digits, that haven’t appeared before. abcd-efg=hij

Or if I use the digits 0-6 inclusive for the initial numbers, the answer can only use the digits 7-9 inclusive?

Correct.
Or algebraically:
abcd – efg = hij

Each digit can only appear once in the overall sum.

Edit: dammit, 26 seconds!

I had bigger things to worry about when I was in Yr5

I think I was still learning my times tables at that age

What’s happening?

alleged cyclists

😆

What’s happening?

We’re running pell mell into a distopian future where ‘facts’ are determined by weight of numbers believing them. But that’s not important right now.

Surely the issue here is an addition not a subtraction (obviously it’s the same thing but easier to “thunk” in terms of addition

Start with two three digit numbers to end with 4…

So 6+4=10
5+3=8
2+7=9

Voila

1089 – 652 = 437

Solve one digit at a time and it’s easy

Anyhow method for a young un…

Maximum sum of two single digits is 17 so the first two digits must sum to 9 or more. (So that when you carry the one you end up with 4 digits. It’s not possible to carry 3 in the right most column so no problems with multiple carries.)

Draw a number square 0-9 on each side. Write in the sums.

Pick numbers which meet your criteria from the number square starting with left most pairs summing to more than 9.

Easy to visualise and follow.

to more than 9.

Should read 9 or more sorry.

Thats a good point.