# Thread: Is B bigger than A?

1. ## Is B bigger than A?

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)

2. ## Re: Is B bigger than A?

Originally Posted by Phairero
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)
... But natural no.s start from 1

3. ## Re: Is B bigger than A?

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!
Can you help me now?

4. ## Re: Is B bigger than A?

Our equation is $191A=70B+898C$

Part 1)

The equation can be rearranged to form $-898C=70B-191A$

We know that $0\leqslant C \leqslant 9$ (C is positive, so a positive no times a negative no is negative), so LHS is negative & $B$ and $A$ are positive so this leads us to the fact that $191A>70B$.

Part 2)

The equation can be rearranged to form $-70B=898C-191A$

Again here, $191A>898C \Leftrightarrow A>C$

5. ## Re: Is B bigger than A?

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.

6. ## Re: Is B bigger than A?

You may or may not say that $B>A$ because we found in part 1 that $191A>70B$. This does not tell us whether $A>B$ or $B>A$.

In part 2 we found that $191A>898C$ but here $898>191$ so we can easily say that $A>C$.

7. ## Re: Is B bigger than A?

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.

8. ## Re: Is B bigger than A?

Hello, Phairero!

Solve the alphametic: . $1998\cdot ABC\,=\,8991\cdot C\!B\!A$

$ABC$ = 3-digit number
$C\!B\!A$ = same 3 digit number written backwards
$A,B,C$ are digits.

We have: . . . $1998\cdot ABC \:=\: 8991\cdot C\!B\!A$

Divide by 999: w $2\cdot ABC \:=\:9\cdot C\!B\!A$

. . . . $2(100A + 10B + 1C) = 9(100C + 10B + 1A)$

. . . . . . $200A + 20B + 2C = 900C + 90B + 9A$

. n . . . . . . . . . . . . $191A \:=\:70B + 898C$ .[1]

. n . . . . . . . . . . . . $191A \:\equiv\: 898C\text{ (mod 70)}$

. . . . . . . . . . . . . . . $51A \:\equiv\:58C\text{ (mod 70)}$

Multiply by 11:. . . . $561A \:\equiv\:638C\text{ (mod 70)}$

. w . . . . . . . . . . . . . . $A \:\equiv\: 8C\text{ (mod 70)}$

Since $A$ and $C$ are digits: . $A \,=\,8,\;C\,=\,1$

Substitute into [1]: . $191(8) = 70B + 898(1)$

. . Hence: . $B \,=\,9$

Therefore: . $\boxed{ABC \:=\:891}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Check:. . $\begin{Bmatrix}1998 \times 891 \\ 8991 \times 198 \end{Bmatrix} \;=\;1,\!780,\!218$

9. ## Re: Is B bigger than A?

I did some research, and understood most of it, except how you got from
$561A \:\equiv\:638C\text{ (mod 70)}$
to
$A \:\equiv\: 8C\text{ (mod 70)}$

EDIT :

And from this

$191A \:\equiv\: 898C\text{ (mod 70)}$
to
$51A \:\equiv\:58C\text{ (mod 70)}$

or can I just write them there and every teacher/tutor/professor/smarter-person-than-me would understand me?