## THINK + RETHINK = REFLECT (each letter stands for a number)

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little_dragon
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### THINK + RETHINK = REFLECT (each letter stands for a number)

I need to find the numbers for each letter in "THINK + RETHINK = REFLECT". They gave me that "I" stands for the number "1".

Since the "RE" is the same on the second number and the answer, I really only have to have the numbers for "THINK + THINK = FLECT". Since T + T = F and no carry, and T can't be 0, then T = 1, 2, 3, 4. Since K + K = T or T + 10, then T has to be even. Then T = 2 or 4.

Am I doing this the right way? Is there a faster way of doing this? Thanks.

stapel_eliz
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little_dragon wrote:Am I doing this the right way? Is there a faster way of doing this? Thanks.

You're doing this exactly the right way. No, I'm not familiar with any "faster" way. Just keep going like you are.

You have found that T must be 2 or 4. Each of these options is a "case". So test the "cases". If T = 2 and I = 1, then... what happens? And so forth. Give yourself plenty of time, and work neatly. If you keep track of all of your cases, you will find an answer!

Eliz.

DAiv
Posts: 35
Joined: Tue Dec 16, 2008 7:47 pm
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### Re: THINK + RETHINK = REFLECT (each letter stands for a number)

little_dragon wrote:I need to find the numbers for each letter in "THINK + RETHINK = REFLECT".

Bear in mind that there may be more than one solution, as some numbers may swap places without affecting any other numbers.

This can occur when you have two sets of adjacent (side-by-side) columns that behave the same with regards to carry overs.

For instance,

$\begin{array}
& BARK\\
+& BARK\\
\hline\\
& YELP\\
\end{array}$

... could lead to two equally valid solutions...

$\begin{array}
& 1927\\
+& 1927\\
\hline\\
& 3854\\
\end{array}
\hspace{30}
\begin{array}
& 1729\\
+& 1729\\
\hline\\
& 3458\\
\end{array}$

The 9s column causes a carry over to the next column to the left, just as the 7s column does. Also, none of the numbers in the result clash with those in the operands (the values we're adding together). So, the 7s column and the 9s column can be freely interchanged without upsetting the validity of the sum.

We must be careful, however, as the following example may at first seem to indicate interchangeability of columns, but on closer inspection, it does not.

$\begin{array}
& 3927\\
+& 3927\\
\hline\\
& 7854\\
\end{array}
\hspace{30}
\begin{array}
& 3729\\
+& 3729\\
\hline\\
& 7458\\
\end{array}$

If we look at the pattern of the numbers, looking for like and unlike digits, we see that the patterns are not the same.

$\begin{array}
& \~9\~\~\\
+& \~9\~\~\\
\hline\\
& 7\~\~\~\\
\end{array}
\hspace{30}
\begin{array}
& \~7\~\~\\
+& \~7\~\~\\
\hline\\
& 7\~\~\~\\
\end{array}$

In the first sum, the numbers above and below the line are different, but in the second, all the numbers are the same. Since each unique digit represents a unique letter in the original puzzle, the second pattern does not match up and so this solution must be wrong.

Well, I hope this insight may prove useful.

Happy puzzling!

DAiv