Re: OT: Interesting math problem...
 

Unless this book is reknowned for giving trick questions, this is most likely not the solution.

Re: OT: Interesting math problem...
 

That is what I am thinking. As far as I know the book is just a regular math book for his grade. Unfortunately they cut the answers out of the back... http://forum.shrapnelgames.com/images/smilies/frown.gif http://forum.shrapnelgames.com/images/smilies/laugh.gif

Re: OT: Interesting math problem...
 

Fyron, that's why I quickly deleted my post. But not quickly enough.

I remeber that I saw such tasks in a school. Common way to solve them is to rewrite example to:

10000*T+1000*R+100*I+10*E+D+10000*D+1000*R+100*I+1 0*V+E-10000*R-1000*I-100*V-10*E-T=0

With additional condition it should be quite easy to solve.

Re: OT: Interesting math problem...
 

Got it:

17465 +
57496 =
-------
74961

T=1
R=7
I=4
E=6
D=5
V=9

I brute forced it, not having time for fancy shmancy formulae. (so, I could be wrong, very wrong!)

Re: OT: Interesting math problem...
 

This problem is the proof I no longer know to do additions; while I figured out the V=9 part, I didn't think that I+I=4 could also mean I=7, and not only I=2. http://forum.shrapnelgames.com/image...s/rolleyes.gif

Re: OT: Interesting math problem...
 

Yay Spoon. I was getting ready to post that it was unsolvable. http://forum.shrapnelgames.com/images/smilies/happy.gif I had worked out that V was 9 and I was 4. But I got hung up thinking then that the only option for R was 2.

Re: OT: Interesting math problem...
 

Yay for spoon! Thanks a lot man! I really do appreciate it. Go have a few brewski's at the Cantina on my tab. http://forum.shrapnelgames.com/images/smilies/laugh.gif

Re: OT: Interesting math problem...
 

It looks like is correct
17465 +
57496 =
_____
74961

Brute force solving (sort of)
Assumptions are that all letters represent different numbers and standard notation is used (the first number in a string is not zero)
V must be either 0 (if D+E < 10) or 9 (if D+E > 9)
If V is 0 then I is 5; if V is 9 then I is 4
For either value of V, R must be either 2 or 7; but R can't be 2 since that would require both T and D to be 1
So R is 7
Then T+D = 6 I tried the 4 possibilities and the only one that worked is the form that I gave above.

Re: OT: Interesting math problem...
 

A more rigorous "proof", took me about five minutes before I went to class, posting it now after class:

<font class="small">Code:</font><hr /><pre> TRIED
+DRIVE
------
RIVET

Assuming we disregard the trivial solution (all zeros), and
force each letter to be distinct (ie for any x and y in
{T,R,I,E,D,V}, the values x != y), and noticing that the
carry can be at most 1 (9 + 9 + 1 = 19):

either E + V = E (V = 0)
or E + V + 1 = E (V = 9)
disregard V = 0 as uninteresting, so V = 9
and D + E &gt; 10 and T &lt; D, T &lt; E
either I + I = V = 9
or I + I + 1 = V = 9
I + I != 9, so I + I + 1 = V is true
so I = {4,9}, but V = 9, so I = 4
so R + R = I = 4 (no carry from previous digit)
then R = {2,7}
V = 9, I = 4, R = {2,7}
either T + D = R = 2
or T + D + 1 = R = 7
assume there is no final carry
so if T + D = 2, then T = D = 1
then T + D + 1 = 7 = R
then T &lt; (7 - D)
V = 9, I = 4, R = 7, T &lt; (7 - D), T &lt; E
try T = 1:
V = 9, I = 4, R = 7, T = 1, D = 5, E = 6</pre><hr />

Re: OT: Interesting math problem...
 

Here is a tough one:
Fill in the long division using the blanks and the given numbers.
<font class="small">Code:</font><hr /><pre>
_ _ 8 _
_______________
_ _ _ | _ _ _ _ _ _ 5
_ _ _ _
========
_ _ _ _
_ _ _
========
_ _ _ _
_ _ _
========
_ _ _ _
_ _ _ _
========
0
</pre><hr />