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05 Apr 2006, 08:53
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K is a two-digit number, if the sum of the tens and unit digits is 3 less than the product of the two digits. What is the value of K?
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05 Apr 2006, 09:12
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K is a two-digit number, if the sum of the tens and unit digits is 3 less than the product of the two digits. What is the value of K?
Let's set the tens digit as x and the units digit as y.
X+Y+3 = X*Y
I think that this then requires a slight bit of logic. X and Y need to be small because otherwise their product would be too great. After considering a couple of values, I can say that K is 33.
Is their a "foolproof" method to solving that equation? I realized that the equation could be set as x - xy + y = -3, but couldn't remember how to solve for x and y as a quadratic. Any thoughts?
Let's set the tens digit as x and the units digit as y.
X+Y+3 = X*Y
I think that this then requires a slight bit of logic. X and Y need to be small because otherwise their product would be too great. After considering a couple of values, I can say that K is 33.
Is their a "foolproof" method to solving that equation? I realized that the equation could be set as x - xy + y = -3, but couldn't remember how to solve for x and y as a quadratic. Any thoughts?
05 Apr 2006, 09:34
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K = 10A + B
A + B = A*B-3
A*B - A = B + 3
A*(B - 1) = B + 3
A = (B + 3)/(B - 1)
Then check the possible cases where both A and B
are integers. Start with B = 0 and end with B = 9.
This leads to 52, 33 and 25, so K cannot be determined.
A + B = A*B-3
A*B - A = B + 3
A*(B - 1) = B + 3
A = (B + 3)/(B - 1)
Then check the possible cases where both A and B
are integers. Start with B = 0 and end with B = 9.
This leads to 52, 33 and 25, so K cannot be determined.
