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andromeda
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  andromeda said:
A, B and X are digits (0... 9) forming numbers, so you have to guess the first digit of the number XAB!

X=?

Have fun! ;)

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A=2

B=0

X=4

20 X 20 = 420 - 20

I'm not sure if I did this right..

Edit: Beckie beat me to it. :o

Edited by Jane
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Maybe I am wrong, but I assumed that A, B, and X were unique (no two were equal), which means that X=1 would not be a valid answer. However, since that condition wasn't stated, I can't say it is wrong.

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  shoeshine said:
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for A=1...3 and B=0, X=A^2. Should X be able to carry two digits then for A=1...9 and B=0, X=A^2

Are there any other solutions?

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No other solutions other than X= 1,4 and 9

since AB * AB = XAB - AB

we can write AB as 10A + B

and XAB as 100X + 10A + B

so (10A+B)2 = 100X+10+B - (10A+B)

or 100A2 + B2 + 20AB = 100X

or X = A2 + B2/100 + AB/5

since X is a number less than 10,

(B2 + 20AB) = 100n for integer values of n

the above equation is statisified only for n=0 and that means B=0

so X= A2

and since X<10 X can take values of 1, 4 and 9

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  vimil said:
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x can be 1,4 or 9

10*10 = 110 - 10

20*20 = 420 - 20

30*30 = 930 - 30

Actually

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I think that x could be number 1, but since we have XAB and not XXB, you know what I mean. So only two solutions!

20*20 = 420 - 20

30*30 = 930 - 30

Nice job! ;)

Edited by andromeda
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  beckie-x said:
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X = 4

20 x 20 = 420 - 20

Kudos to MaestroOD.

You bad girl! Hm... the next one will be much harder! We'll see how MaestroOD copes with that one ;)

There are

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two solutions for this one BTW!

:P

Edited by andromeda
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For A, B and X are digits (0... 9)

AB X AB = XAB - AB

1) A = 0, B = 1, X=0,1,2,3,4,...,9 infact X can be anything

2) A = 1, B = 2, X = 3

3) A = 2, B = 3, X = 7

4) A = 1, B = 3, X = 4

5) A = 1, B = 4, X = 5

6) A = 1, B = 5, X = 6

7) A = 1, B = 6, X = 7

8) A = 1, B = 7, X = 8

9) A = 1, B = 8, X = 9

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