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
Part 1)
The equation can be rearranged to form
We know that (C is positive, so a positive no times a negative no is negative), so LHS is negative & and are positive so this leads us to the fact that .
Part 2)
The equation can be rearranged to form
Again here,
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.
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: .
= 3-digit number
= same 3 digit number written backwards
are digits.
We have: . . .
Divide by 999: w
. . . .
. . . . . .
. n . . . . . . . . . . . . .[1]
. n . . . . . . . . . . . .
. . . . . . . . . . . . . . .
Multiply by 11:. . . .
. w . . . . . . . . . . . . . .
Since and are digits: .
Substitute into [1]: .
. . Hence: .
Therefore: .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Check:. .
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?