I know that
| 191*A = 70*B + 898*C
| 70*B = 191*A - 898*C
| 898*C = 191*A - 70*B
I can say for sure that C<A
But can I say B>A without guessing ?
(A,B,C are natural numbers 0-9)
Well, some include 0 too.
What I meant was they are ... numbers from 0-9
as in (0,1,2,3,4,5,6,7,8,9)
(Edit (added) : NOT 1,3 or pi or 3,89 or something else like that)
What were they called... oh, one-digit numbers!
Sorry my bad.
Can you help me now?
Our equation is $\displaystyle 191A=70B+898C$
Part 1)
The equation can be rearranged to form $\displaystyle -898C=70B-191A$
We know that $\displaystyle 0\leqslant C \leqslant 9$ (C is positive, so a positive no times a negative no is negative), so LHS is negative & $\displaystyle B $ and $\displaystyle A$ are positive so this leads us to the fact that $\displaystyle 191A>70B$.
Part 2)
The equation can be rearranged to form $\displaystyle -70B=898C-191A$
Again here, $\displaystyle 191A>898C \Leftrightarrow A>C$
Yes, I apologize If I can't read it out, but where does it say then, that B > A ?
Can I say it?
I already know the answer I just need to know if I can say that B is bigger than A
If I can't say it my whole solution might just fall apart.
You may or may not say that $\displaystyle B>A$ because we found in part 1 that $\displaystyle 191A>70B$. This does not tell us whether $\displaystyle A>B$ or $\displaystyle B>A$.
In part 2 we found that $\displaystyle 191A>898C$ but here $\displaystyle 898>191$ so we can easily say that $\displaystyle A>C$.
That is sad...
Then how am I supposed to figure this out
1998*ABC=CBA*8991
ABC = 3 digit number
CBA = same 3 digit number written backwards
A,B,C are one digit numbers.
What I did until that.
1998 * ABC = CBA * 8991
1998 * (100A + 10B + 1C) = (100C + 10B + 1A) * 8991
199800A + 19980B + 1998C = 899100C + 89910B + 8991A
199800A - 8991A = 89910B - 19980B + 899100C - 1998C
190809A = 69930B + 897102C |divided by 9
21201A = 7770B + 99678C |divided by 3
7067A = 2590B + 33226C |divided by 37
191A = 70B + 898C
191 is a prime number, so I can't divide any further.
Hello, Phairero!
Solve the alphametic: .$\displaystyle 1998\cdot ABC\,=\,8991\cdot C\!B\!A$
$\displaystyle ABC$ = 3-digit number
$\displaystyle C\!B\!A$ = same 3 digit number written backwards
$\displaystyle A,B,C$ are digits.
We have: . . .$\displaystyle 1998\cdot ABC \:=\: 8991\cdot C\!B\!A$
Divide by 999: w$\displaystyle 2\cdot ABC \:=\:9\cdot C\!B\!A$
. . . .$\displaystyle 2(100A + 10B + 1C) = 9(100C + 10B + 1A)$
. . . . . . $\displaystyle 200A + 20B + 2C = 900C + 90B + 9A$
. n . . . . . . . . . . . . $\displaystyle 191A \:=\:70B + 898C$ .[1]
. n . . . . . . . . . . . . $\displaystyle 191A \:\equiv\: 898C\text{ (mod 70)}$
. . . . . . . . . . . . . . . $\displaystyle 51A \:\equiv\:58C\text{ (mod 70)}$
Multiply by 11:. . . . $\displaystyle 561A \:\equiv\:638C\text{ (mod 70)}$
. w . . . . . . . . . . . . . . $\displaystyle A \:\equiv\: 8C\text{ (mod 70)}$
Since $\displaystyle A$ and $\displaystyle C$ are digits: .$\displaystyle A \,=\,8,\;C\,=\,1$
Substitute into [1]: .$\displaystyle 191(8) = 70B + 898(1)$
. . Hence: .$\displaystyle B \,=\,9$
Therefore: .$\displaystyle \boxed{ABC \:=\:891}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Check:. . $\displaystyle \begin{Bmatrix}1998 \times 891 \\ 8991 \times 198 \end{Bmatrix} \;=\;1,\!780,\!218$
I did some research, and understood most of it, except how you got from
to
EDIT :
And from this
to
or can I just write them there and every teacher/tutor/professor/smarter-person-than-me would understand me?