Matix problem 2
May 24th, 2014 at 4:42:44 AM
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On behalf of our enthusiastic new member Matix, I post the following problem:
With all ten digits (0,1,2,3,4,5,6,7,8,9), each used once, we made two integers (X and Y with X> Y), their sum and difference.
Find these two numbers.
Good luck !
I have found the solution, but will wait a while before posting since I had a head start...
With all ten digits (0,1,2,3,4,5,6,7,8,9), each used once, we made two integers (X and Y with X> Y), their sum and difference.
Find these two numbers.
Good luck !
I have found the solution, but will wait a while before posting since I had a head start...
May 24th, 2014 at 4:49:45 AM
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Thank you very much ! ;)
May 24th, 2014 at 5:08:19 AM
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Question seems a bit confusing. I understand the sum part, but the difference?
For example, Y=01234, X=56789.
Sum Y = 0+1+2+3+4 = 10
Sum X = 5+6+7+8+9 = 35
So x > y = 35 > 10
What's the difference?
For example, Y=01234, X=56789.
Sum Y = 0+1+2+3+4 = 10
Sum X = 5+6+7+8+9 = 35
So x > y = 35 > 10
What's the difference?
May 24th, 2014 at 5:23:13 AM
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No, X+Y and X-Y...
May 24th, 2014 at 6:04:45 AM
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Maybe I'm not the sharpest tool in the shed, but I still don't understand what is being asked.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 24th, 2014 at 6:31:39 AM
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X
Y
A = X+Y
B = X-Y
X,Y,A,B together contain digits 1,2,3,4,5,6,7,8,9,0 exactly once
example: 456 + 123 = 780; 456 - 123= 9
(at least, that is what I solved...)
Y
A = X+Y
B = X-Y
X,Y,A,B together contain digits 1,2,3,4,5,6,7,8,9,0 exactly once
example: 456 + 123 = 780; 456 - 123= 9
(at least, that is what I solved...)
May 24th, 2014 at 8:05:02 AM
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Exactly !
May 24th, 2014 at 11:53:00 AM
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grace period is over, here is my
X=146
Y=57
X+Y= 203
X-Y= 89
X=146
Y=57
X+Y= 203
X-Y= 89
