yr 5 maths puzzle - help - can only find unstisfactory solution

[Spatz]

Maths puzzle sent home for top maths group in yr 5

In the multiplication below, some of the digits have been replaced by letters and others by asterisks. Where a digit has been replaced by a letter, the same letter is used each time, and different letters replace different digits. Can you reconstruct the original multiplication?

........A B C
........B A C
.....* * * *
.....* * A
* * * B

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(please excuse layout - had to put . to keep columns. Ignore "."s)

Absolutely no way my DD can do this.
Anyone have good solution and explanation for 9 yr old?

So.... C x ABC is a 4 digit number which doesn't contain any of the original digits

A x ABC is a three digit number ending in A

B x ABC is a 4 digit number ending in B

I think I'd start looking at A x ABC and try to work out what A needs to be. Um, A can't be bigger than 3 because 4 x 400 would need 4 digits. So A is 1,2, or 3..... And A x C has to equal A (plus 10s maybe) C could be 1, and A 2 or 3. Maybe get her thinking what else would work?

eg 2 x 6 would be 12 so you could have A=2 and B=6

I'd go through the possibilities there as a next step....

I have gone through these sort of steps, but am unhappy with the solution I've found mostly because I feel all the A, B and C should have been filled in not given asterisks.

Do you mean some of the *s turn out to be the same digit as A,B,or C? I wouldn't like that either! It should be all of them fr consistency.... but maybe they mean you only need the ones they've put in to do the sums? Still, I don't like it either! Not logical....

a=2, c=1? Am I one the right track?

on

But ha! end up with a * that should be a C!!

I agree the last * ends up being one of A,B or C.
This is what's bugging me. Apart from the fact I don't think a 9 yr old would have a clue!!

We got to C=1 together, but that means the first line of the multiplication only has 3 digits.

Some random thoughts.

C x C gives you a number that doesn't end in C.
So that rules out C being 1, 5 or 6.

A x C gives you a number ending in A
B x C gives you a number ending in B

I've drawn myself a 9 x 9 grid of the last digit of the resulting products.

But as C cannot be 1,5 or 6, the only number that A and B can be is 5!!

As they can't both be 5, unless I've done something wrong (probably) it seems like an impossible puzzle.

It is only possible if some of the asterisks are A,B or C.
Which is very very annoying and NOT HOW MATHS SHOULD BE!

If I got a puzzle like that in a puzzle magazine, I would complain about the lack of clarity in the gven rules. I would have assumed all instances of A, B or C should have been marked, since some of them were, or that the instructions said clearly that there might be others unmarked.

ac=a (mod 10)

possible pairs (a,c) are:
a=1, c=1
a=2, c=1,6
a=3, c=1
a=4, c=1,6
a=5, c=1,3,5,7,9
a=6, c=1,6
a=7, c=1
a=8, c=1,6
a=9, c=1
a=0, c=anything

c squared is not c (mod 10)

so c cannot be 0, 1, 5, 6

hence a must be 5 or 0.

b is subject to the same constraints as a, so b must be 5 or 0.

Therefore it is not possible to complete this puzzle using different three-digit numbers. I have a first in maths from Oxford btw.

so DH sits down in front of the computer and sees my scribblings and decides to run a computer programme to find any possible solutions. It turns out there is only one possible solution and this has c squared equal to c.

The answer is a=2, b=8, c=6

286

826 *

--

1716

572

2288 +

--

236236

wait a minute, the formatting was all wrong:

_286

_826 *

--

__1716

__572

2288 +

--

236236

no, formatting still all wrong, but hopefully you see what I mean.

Hurrah for your DH, but I did it on the front of the guardian magazine [smug emoticon]

It is still very annoying!

What shall I say to DD's teacher on Mon?
He is lovely BTW

AAM's DH here. Manual solution (I guess equivalent to your scribblings on leftwing propaganda:

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We may assume A and B are non-zero

Consider C*ABC = a four digit number
: C isn't 0 or 1

Consider A*ABC = a three digit number
: A is 1, 2 or 3

Now consider A*ABC ends in A, remembering that 4>A>0 and C>1
: C=2 => A=0 (bad)
: C=3 => A=0 (bad) or 5 (bad)
: C=4 => A=0 (bad)
: C=5 => A=0 (bad) or 5 (bad)
: C=6 => A=0 (bad) or 2 (good!)
: C=7 => A=0 (bad) or 5 (bad)
: C=8 => A=0 (bad)
: C=9 => A=0 (bad) or 5 (bad)

Wow, so that's given us A=2 and C=6 :) Now what's B?

Consider B*2B6 = a four digit number
: B is 5,7,8 or 9 (note: not 6, because C=6)

Now consider B*2B6 ends in B
: B=5 => no
: B=7 => no
: B=8 => yes!
: B=9 => no

So B=8. That completes the puzzle.

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The missus will be furious when she sees I've posted using her MN account! Hahahaha :)

Thanks AAM's DH

Good luck when she finds out!