Find A, B, C
1998xABC=CBAx8991
I appreaciate if someone could help. I tried some numbers and some stuff but worthless..
Hello, Trolo!
This is obviously an "alphametic",
. . where $\displaystyle ABC$ and $\displaystyle CBA$ represent 3-digit numbers.
$\displaystyle \text{Find }A, B, C .$
. . $\displaystyle 1998 \times ABC \:=\:CBA \times 8991$
I began an orderly (but lengthy) search
. . and virtually tripped over the solution.
The problem has this form:
. . $\displaystyle \begin{array}{ccccccc} &&& 1 & 9 & 9 & 8 \\ && \times && A & B & C \\ && - & - & - & - & - \\ && * & * & * & * & * \\ & * & * & * & * & * \\ * & * & * & * & * \\ - & - & - & - & - & - & - \\ * & * & * & * & * & * & * \end{array} \qquad\begin{array}{ccccccc} &&& 8 & 9 & 9 & 1 \\ && \times && C & B & A \\ && - & - & - & - & - \\ && * & * & * & * & * \\ & * & * & * & * & * \\ * & * & * & * & * \\ - & - & - & - & - & - & - \\ * & * & * & * & * & * & * \end{array}$
Suppose $\displaystyle C = 1$, then $\displaystyle A = 8.$
Then we have:
. . $\displaystyle \begin{array}{ccccccc} &&& 1 & 9 & 9 & 8 \\ && \times && 8 & B & 1 \\ && - & - & - & - & - \\ && & 1 & 9 & 9 & 8 \\ & * & * & * & * & * \\ 1 & 5 & 9 & 8 & 4 \\ - & - & - & - & - & - & - \\ * & * & * & * & * & * & 8 \end{array} \qquad\begin{array}{ccccccc} &&& 8 & 9 & 9 & 1 \\ && \times && 1 & B & 8 \\ && - & - & - & - & - \\ && 7 & 1 & 9 & 2 & 8 \\ & * & * & * & * & * \\ & 8 & 9 & 9 & 1 \\ - & - & - & - & - & - & - \\ * & * & * & * & * & * & 8 \end{array}$
The suspicious "shape" of the two factors: .$\displaystyle 1\,9\,9\,8\,\text{ and }\,8\,B\,1$
. . led me to try $\displaystyle B = 9$ first . . . and it worked!
. . . . . $\displaystyle 1998 \times 891 \;=\;8991 \times 198 \;=\;1,\!780,\!218$
Therefore: .$\displaystyle \begin{Bmatrix}A \:=\:8 \\ B \:=\:9 \\ C\:=\:1\end{Bmatrix}$
Multiplication is commutative, which means that the order you multiply is unimportant - you'll get the same result regardless.
Therefore, ABC = BCA = CAB = CBA
If you divide any number by itself, you get 1:
$\displaystyle \frac {x^1}{x^1} = x^1^-^1 = x^0 = 1$
So,
$\displaystyle 1998*ABC=CBA*8991$
$\displaystyle \frac {1998 * ABC}{CBA} = 8991$
ABC and CBA cancel out, to give:
$\displaystyle 1998 * \frac{1}{1} = 8991$
$\displaystyle 1998 = 8991$
Which, of course, makes no sense, so ABC must equal 0. They must all be zero.
EDIT: Sorry, I missed your post and assumed they were variables... my mistake! I'll leave this here anyway.