1. ## Thank you!

Originally Posted by JeWiSh
a=7
b=6
c=5
d=2
e=3
f=4

I did it with some brief trial and error, and I'm not sure if maybe there's a more efficient, concrete method...
That is brilliant. Thank you so much. I figured it was possible with trial and error but I guess I am a bit tired/out of practice.

Thanks again

2. ## Finding algebra variables

Hi,

Trying to help daughter with homework and it has been so long! Can't get my head round this one:

ab*c+d=200+ef

We are told that a,b,c,d,e,f represent 2,3,4,5,6,7 but not in that order. Can someone tell me how to work this out please, it is driving me nuts. Thank you very much.

3. Originally Posted by myfi
Hi,

Trying to help daughter with homework and it has been so long! Can't get my head round this one:

ab*c+d=200+ef

We are told that a,b,c,d,e,f represent 2,3,4,5,6,7 but not in that order. Can someone tell me how to work this out please, it is driving me nuts. Thank you very much.
a=7
b=6
c=5
d=2
e=3
f=4

I did it with some brief trial and error, and I'm not sure if maybe there's a more efficient, concrete method...

4. Hello, myfi!

Trying to help daughter with homework.
. . $AB*C+D\:=\:200+EF$

We are told that $A,B,C,D,E,F$ represent 2,3,4,5,6,7 but not in that order.

I don't understand that solution
. . . or maybe I don't understand the problem.

I would assume that $AB$ represents a two-digit number, as does $EF.$

Otherwise, why would they write it that way?

So we have a math puzzle to solve:

. . $\begin{array}{ccc}
& A & B \\ & \times & C \\ \hline
? & ? & ? \\
+ & & D \\ \hline
2 & E & F \end{array}$

There are $6! = 720$ possible placements of those digits.

We can eliminate many possibilities,
. . but it will still require Brute Force trial-and-error.

I found a solution (and I'm not looking for any more).

. . $\begin{array}{ccc}
& {\color{red}3} & {\color{red}6} \\ & \times & {\color{red}7} \\ \hline
2 & 5 & 2 \\
+ & & {\color{red}2} \\ \hline
2 & {\color{red}5} & {\color{red}4} \end{array}$