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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- October 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.. - October 10th 2011, 07:20 AMe^(i*pi)Re: Need help
- October 10th 2011, 08:20 AMTroloRe: Need help
Will it change something if i say that A,B,C are natural numbers.

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

This is obviously an "alphametic",

. . where and represent 3-digit numbers.

Quote:

. .

I began an orderly (but lengthy) search

. . and virtuallyover the solution.*tripped*

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: .

- October 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:

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. - October 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