# 1998xABC=CBAx8991: Find A, B, C.

• Oct 10th 2011, 07:13 AM
Trolo
1998xABC=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 AM
e^(i*pi)
Re: Need help
Quote:

Originally Posted by Trolo
Find A, B, C

1998xABC=CBAx8991

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

Order doesn't matter in multiplication so ABC would cancel to give $1998=8991$ which is clearly not the case.

Therefore all we can say is that $ABC = 0$ and individual values cannot be determined
• Oct 10th 2011, 08:20 AM
Trolo
Re: Need help
Will it change something if i say that A,B,C are natural numbers.
• Oct 10th 2011, 09:00 AM
Soroban
Re: Need help
Hello, Trolo!

This is obviously an "alphametic",
. . where $ABC$ and $CBA$ represent 3-digit numbers.

Quote:

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

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

I began an orderly (but lengthy) search
. . and virtually tripped over the solution.

The problem has this form:

. . $\begin{array}{ccccccc} &&& 1 & 9 & 9 & 8 \\ && \times && A & B & C \\ && - & - & - & - & - \\ && * & * & * & * & * \\ & * & * & * & * & * \\ * & * & * & * & * \\ - & - & - & - & - & - & - \\ * & * & * & * & * & * & * \end{array} \qquad\begin{array}{ccccccc} &&& 8 & 9 & 9 & 1 \\ && \times && C & B & A \\ && - & - & - & - & - \\ && * & * & * & * & * \\ & * & * & * & * & * \\ * & * & * & * & * \\ - & - & - & - & - & - & - \\ * & * & * & * & * & * & * \end{array}$

Suppose $C = 1$, then $A = 8.$
Then we have:

. . $\begin{array}{ccccccc} &&& 1 & 9 & 9 & 8 \\ && \times && 8 & B & 1 \\ && - & - & - & - & - \\ && & 1 & 9 & 9 & 8 \\ & * & * & * & * & * \\ 1 & 5 & 9 & 8 & 4 \\ - & - & - & - & - & - & - \\ * & * & * & * & * & * & 8 \end{array} \qquad\begin{array}{ccccccc} &&& 8 & 9 & 9 & 1 \\ && \times && 1 & B & 8 \\ && - & - & - & - & - \\ && 7 & 1 & 9 & 2 & 8 \\ & * & * & * & * & * \\ & 8 & 9 & 9 & 1 \\ - & - & - & - & - & - & - \\ * & * & * & * & * & * & 8 \end{array}$

The suspicious "shape" of the two factors: . $1\,9\,9\,8\,\text{ and }\,8\,B\,1$
. . led me to try $B = 9$ first . . . and it worked!

. . . . . $1998 \times 891 \;=\;8991 \times 198 \;=\;1,\!780,\!218$

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

• Oct 10th 2011, 09:03 AM
FrameOfMind
Re: 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:

$\frac {x^1}{x^1} = x^1^-^1 = x^0 = 1$

So,
$1998*ABC=CBA*8991$
$\frac {1998 * ABC}{CBA} = 8991$
ABC and CBA cancel out, to give:
$1998 * \frac{1}{1} = 8991$
$1998 = 8991$
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 AM
Wilmer
Re: 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