# Division Question

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• March 24th 2010, 04:15 AM
nazz
Division Question
How many different solutions are there to this division? different letters stand for different digits and no number begins with zero.

JMO / UK = OK ( what are the numbers)
• March 24th 2010, 04:56 AM
sa-ri-ga-ma
Quote:

Originally Posted by nazz
How many different solutions are there to this division? different letters stand for different digits and no number begins with zero.

JMO / UK = OK ( what are the numbers)

One solution.
JMO = OK*UK
K cannot be one, because K^2 = 1 = O,. repetition is not allowed.
Let K = 2. Then O = 4
So 42* U2 = JMO.
Product is three digits. U cannot be 3,neither 2. So try one
So 42*12 = 504.
Try other combination, if possible.
• March 24th 2010, 06:24 AM
Grandad
Hello nazz
Quote:

Originally Posted by nazz
How many different solutions are there to this division? different letters stand for different digits and no number begins with zero.

JMO / UK = OK ( what are the numbers)

I think the complete set of solutions is:
$\begin{array}{c c c c c}
J&M&O&U&K\\
5&0&4&1&2\\
8&9&6&1&4\\
7&4&1&3&9\\
9&3&1&4&9\\
\end{array}$
Grandad
• March 24th 2010, 07:14 AM
sa-ri-ga-ma
Quote:

Originally Posted by Grandad
Hello nazzI think the complete set of solutions is:
$\begin{array}{c c c c c}
J&M&O&U&K\\
5&0&4&1&2\\
8&9&6&1&4\\
7&4&1&3&9\\
9&3&1&4&9\\
\end{array}$
Grandad

In the last set there is a repetition which is not allowed.
• March 24th 2010, 07:37 AM
Grandad
Quote:

Originally Posted by sa-ri-ga-ma
In the last set there is a repetition which is not allowed.

This is true! Strike that line from the solutions!

Grandad
• March 24th 2010, 08:32 AM
nazz
thanks for all the replies, is there a way solving it by equations, instead of solving it by trials