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K is a two-digit number, if the sum of the tens and unit

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SVP
Joined: 24 Sep 2005
Posts: 1884

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K is a two-digit number, if the sum of the tens and unit [#permalink]

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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?

Kudos [?]: 379 [0], given: 0

Manager
Joined: 09 Feb 2006
Posts: 129

Kudos [?]: 8 [0], given: 0

Location: New York, NY

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05 Apr 2006, 09:12
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?

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Manager
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05 Apr 2006, 09:34
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.

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Manager
Joined: 24 Oct 2005
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05 Apr 2006, 10:22
I agree...k cannot be determined without further information.

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VP
Joined: 29 Dec 2005
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05 Apr 2006, 12:27
yah, there are more than one solution, Laxi.

laxieqv wrote:
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?

how about -30 [10(-3) + 0], since 3 is -ve and 0 cannot be -ve or +ve.

sum = -3
product = 0
diff = -3

though it is only an additional one but i am not sure how correct it is?

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Senior Manager
Joined: 11 Nov 2005
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06 Apr 2006, 13:52
when K has several values,
given in the choice one can select the right answer.

ccax wrote:
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.

Kudos [?]: 16 [0], given: 0

06 Apr 2006, 13:52
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