Find A, B, C

1998xABC=CBAx8991

I appreaciate if someone could help. I tried some numbers and some stuff but worthless..

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- Oct 10th 2011, 07:13 AMTrolo1998xABC=CBAx8991: Find A, B, C.
Find A, B, C

1998xABC=CBAx8991

I appreaciate if someone could help. I tried some numbers and some stuff but worthless.. - Oct 10th 2011, 07:20 AMe^(i*pi)Re: Need help
- Oct 10th 2011, 08:20 AMTroloRe: Need help
Will it change something if i say that A,B,C are natural numbers.

- Oct 10th 2011, 09:00 AMSorobanRe: Need help
Hello, Trolo!

This is obviously an "alphametic",

. . where $\displaystyle ABC$ and $\displaystyle CBA$ represent 3-digit numbers.

Quote:

$\displaystyle \text{Find }A, B, C .$

. . $\displaystyle 1998 \times ABC \:=\:CBA \times 8991$

I began an orderly (but lengthy) search

. . and virtuallyover the solution.*tripped*

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: .$\displaystyle \begin{Bmatrix}A \:=\:8 \\ B \:=\:9 \\ C\:=\:1\end{Bmatrix}$

- Oct 10th 2011, 09:03 AMFrameOfMindRe: Need help
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. - Oct 10th 2011, 10:31 AMWilmerRe: Need help
Soroban, "search" needs not be "lengthy"; since 1998/8991 = 2/7, then:

191A - 898C = 70B : B = (191A - 898C) / 70

a few manipulations lead to A=8, C=1 ; so B=9