Thread: 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..

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 which is clearly not the case.

Therefore all we can say is that and individual values cannot be determined

Will it change something if i say that A,B,C are natural numbers.

Hello, Trolo!

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

. .

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


The problem has this form:

. .


Suppose , then
Then we have:

. .


The suspicious "shape" of the two factors: .
. . led me to try first . . . and it worked!

. . . . .


Therefore: .

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:



So,


ABC and CBA cancel out, to give:


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.

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