Y9 maths homework...

[Erebus]

Though it could be any year!

You have the six numbers, 1,2,3,4,5,6
You have to rearrange them (using them only once apiece) to make one three digit number plus a second three digit number equal a given answer, e.g.

876 (answer) = 541 + 326

The questions are:
1086 = nnn + nnn
381 = nnn + nnn

Is there a methodology or is it just trial and error?

I think it would be refined trial and error (ie you can work out combinations of digits to make the end digit in each case etc).

Might give it a go - could do with stretching my brain a bit this afternoon

Bit of both really. I'd start from the RHS. Consider the 381. The only way to get the '1' at the end is with 5 and 6. So you've got a '1' carried over into the tens. That means you need 7. You've already used 5 and 6, so it will have to be 3 and 4 in the tens. That leaves 2 and 1 in the hundreds. So one possibility is 246+135, but you could interchange the digits between the two eg. 135 and 146.

Does that help?

Done 381 :)

Am stuck with 1,086

245 + 136 = 381

I don't think the other is possible because
6 can be made from 2+4 or 5+l
8 could be 5+3 or 2+6
10 must be 6+4
So no combination of two 3 figure numbers containing only the digits l23456 With no repeats can be added to obtain 1086

(happy to be proved wrong!)

952 + 134

Methodology same as irregular suggested - just look at the number combinations that give 10, 8 and 6.

But you can't use 9 ? (unless I'm missing something)

Oops - just seen that I did not read the OP properly and qu only allows 6 numbers. Sorry!

Sorry, DS tells me it's 1083

thanks for the 381 explanation!

Method for 1083 , is just the same then.
Easy :)

I'm glad it was wrong!

651+432 or 452 +631, it is actually not difficult, just need thinking cap on!

Yes, I have to say it was obvious once we twigged about working back from the end .

I was just about to post that there is no way 1082 works .

1086*