To ask for help with DS maths homework

[Mumski45]

Even though I have a maths degree!
He is 10 and in year 6. I can find a 'cryptarithm solver' using google but need to explain strategy to him so that doesn't help!

This is hurting my brain

I am assuming that there is a 'key' to unlocking it.
Not sure if both should be solved together or not.
In addition to the g=2j I got that A-1=F

Just shown this to DS (17) and a whizz at maths.
His response " I could do it but I don't want to as it would take ages. Ridiculous for Y6!"

I'd be inclined to write a note saying you tried, spent x amount of time on it and are stumped.

DD once got a worksheet in Y6 and she, DH and I spent ages on one question because we could've get it to work. Sent it back and I spoke to her teacher (I worked in the school). She got her DH to look it over (Head of maths at a secondary school) and it wasn't possible!

I did it!!! Do you want the solution?

I've just shown it to my 17yo DD who is also a whizz at maths ...

The 0 must be some thing which is on the top line only I think?

Well, yes - how?

Thanks for your help every one. It's nuts isn't it. I will be speaking to DS teacher and I know I won't be the only one. I just thought that there might be an obvious trick that I was missing?
This was on the same page as a set of standard subtractions which were very doable and I just couldn't understand for the life of me the connection between questions 1-3 and this.

Camefromthedarkside it's not so much the solution as I found a cryptarithm solver on internet which gave me about 10 possible correct solutions. It's more about a strategy to solve it in the amount of time that you would expect a 10 year old to do their homework in.

I used trial and error and a bit of logic. Lots of scribbling too! I'm not sure if there's a more obvious way of solving it. A = f + 1, j = I + 1.

I'm not convinced theres enough info to solve analytically.

DD now wants a maths puzzle which is soluble

I also worked across both calculations at the same time checking the logic of each number. Other than that, I'm not sure there is a quicker method that would work.

Camefromthedarkside are your scribble legible? Would it be something you could take a picture of and post pls. Would be very much appreciated although as was mentioned further up the thread I'm not sure what the learning outcome for DS is :(

It's probably not a huge amount of help as I rubbed out each number as I did it. I think I started with j i and g from the second calculation.

I'd be inclined to write a note saying you tried, spent x amount of time on it and are stumped

This. Homework should be doable.

I have so far got G = 2J
So if G was 8, J = 4
Or G = 6 and J = 3

I wonder about the learning outcome too. I've seen subtraction like this with a few missing digits which reinforces place value, but thus particular one does seem to rely a lot on trial and error, unless I'm missing something

Thanks very much camefromthedarkside

Excuse my crappy typing

You're welcome - I did get a bit too excited when I realised I'd solved it. I had a lucky guess with my first number I think.

It could be used as an exercise in trial and error, lots of equations have been solved like this and then work backwards.

Or it might be an exercise in seeing how different numbers interact with one another, i.e. J=I+1 etc, the top row is 10,000A +1000B + 100C etc

Also quite a good exercise in deducing the certain letters can't be something i.e. A, C, F,G, I, J can't be 0 like sudoku.

Here's an example from you tube of a similar puzzle (but an addition).
m.youtube.com/watch?v=rlwp_LGfHTM
It blew my mind a bit but I'm determined to solve it now! So much for my Saturday night!

Ok I have solved the second one and assume if I plug in the figures it will solve the first.

But I could be wrong so let me check!

Hope this makes sense.

J=2 and I=1 from the second sum first column. As when 1 is "borrowed" from the 2 it will become 1-1 and equal 0.

Therefore G=4 from the second sum last column G-J=J becomes 4-2=2

From the first column of the first sum we know that 1 must be "borrowed" from the A as it becomes F in the answer. So A-1=F

From the second column of the first sum B-F=4 but we also know that B is a smaller number than F as it needs to borrow 1 from A.

A F possible combinations are: 6, 5 7,6 8,7 9,8

So as B-F=4. B can be 9 0 1 2

It can't be 9 as it has to be smaller than F or 1 or 2 as these are already assigned so

B=0

From there F=6 second column first sum

A=7 first column first sum

C=8 second column second sum

E=9 last column first sum

D=3