OT: Interesting math problem...
 

So my cousin has a math problem that none of us can figure out...Here on the forums I am sure someone can figure it out within like say 30 seconds so I will time you...:D

But here is the problem:

TRIED
+DRIVE
------
=RIVET

Each letter represents a number from 0-9. That is pretty much all the help/hint they give in my cousins book for homework as well. I believe this is 9th grade stuff.

But if someone could help me out on this and/or explain how best to go about answering this problem that would be great.

Re: OT: Interesting math problem...
 

*Slaps self* Nevermind; I mixed up the Last two sums, and came to the wrong conclusion.

Re: OT: Interesting math problem...
 

Heck, I barely passed algebra, even in the 9th grade. Can't help you.

Re: OT: Interesting math problem...
 

The only solution I have been able to find is every letter is equal to 0. It is technically a solution; 00000+00000=00000.

Can two letters (or more) share the same value? Do you know if the final result has five figures, or could the = be a sixth number? (Sounds unlikely, but hey, who knows?)

Re: OT: Interesting math problem...
 

Not much time, but it looks like:

V = 0
I = 5

Re: OT: Interesting math problem... *DELETED*
 

Post deleted by aiken

Re: OT: Interesting math problem...
 

well if e = 0

does not matter what the other letters are

Re: OT: Interesting math problem...
 

Erm... that would make V = 0, and there would be a 1 carried to the next column.

Re: OT: Interesting math problem...
 

V=0 was my first attempt, but only works if you have T=D=1 and E=V=0, which may or may not fall under the guideline.

If you can have letters sharing the same number, then all letters should be 0; much easier and faster.

Re: OT: Interesting math problem...
 

The thing is the book he had didn't indicate if numbers could share the same value. It would be nice if they could as everyone said 00000+00000=00000 is a simple solution.

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 />

Re: OT: Interesting math problem...
 

Will, that's a good thourough proof there, but I'm curious why you "discount V=0 as uninteresting". Maybe that's a math thing I don't get, but it seems V=0 is just as valid a posibility as V=9 until it's proven or disproven. In fact I did disprove V=0 in this case, in a very similer fasion to how you proved V=9.

If V=0 then

D + E &lt; 10, since no number for E + 0 + 1 = E or 10 + E
Since D + E &lt; 10, there is no carry over to the second or third columns.
Therefore I = 5, since I must be either 0 or 5 and we have already determined V = 0.
I + I gives us a carry to the next position so R + R + 1 = 5. For this to be correct R must be either 2 or 7. (Somehow I got this point while disproving V = 0, but forgot it while attempting to prove V = 9 ) If R is 2 then there is no carry and T + D is 2. For this to be correct either T and D must both be 1 or T and D are 0 and 2. Both violate our assumptions so R cannot be 2, therefore R = 7.
Therefore 1 + T + D = 7, T + D = 6. T and D cannot both be 3, and neither T or D can be 0(+6) or 5(+1), so T &amp; D must be 2 &amp; 4. To determine which is which we can try and substitute that into D + E = T (Which as we have already calculated is &lt; 10)
If D is 2, E would have to also be 2 for T to be 4, so D cannot be 2.
If D is 4, E would have to be -2 for T to be 2, so D cannot be 4.

Since D cannot be 2 or 4 then it follows that V cannot be 0

Heh, reading this out loud I realized I sound like Vizzini from the Princess Bride. http://forum.shrapnelgames.com/images/smilies/happy.gif

Re: OT: Interesting math problem...
 

Took me about 5 minutes http://forum.shrapnelgames.com/images/smilies/cool.gif
<font class="small">Code:</font><hr /><pre>
9 8 8 9
_______________
1 1 5 | 1 1 3 7 2 3 5
1 0 3 5
========
1 0 2 2
9 2 0
========
1 0 2 3
9 2 0
========
1 0 3 5
1 0 3 5
========
0
</pre><hr />

Re: OT: Interesting math problem...
 

I cannot even understand SJ's question. Am I supposed to be able to comprehend the answer? *Goes back to playing with his abacus*

Re: OT: Interesting math problem...
 

Well, I didn't really elaborate it very well. You have to look ahead a bit to see that V = 0 leads to the trivial (non-distinct) solution of 00000 + 00000 = 00000. While it's not immediately obvious that this is the case, since I don't know of a proof that would be less than your about 9 step proof, there is also a sort of mathematical intuition there, that there already is a trivial solution for zeros, and a non-trivial solution most likely will not have the zeros. But, overall, I was just looking to solve it quickly, elegantly, and I determined that rather than proving V != 0, and doing most of the problem twice, I would just ignore that possibility, and go with the more "interesting" one. And, usually, that works http://forum.shrapnelgames.com/images/smilies/happy.gif

Re: OT: Interesting math problem...
 

Does one of you math geniuses know what the equation for finding the square root of something is?

Re: OT: Interesting math problem...
 

It is rather complicated. This site might prove of some interest. Just found it on Google, but looks interesting... http://www.qnet.fi/abehr/Achim/Calcu...ier_rods2.html

Re: OT: Interesting math problem...
 

That's not an equation, it's a, well, I'm not sure what to call it. A human program in english? My math knowledge says an equation is impossible, by the way, because with an equation you're trying to find a) the number that was multiplied and b) the number it was multiplied by(Please note that while these numbers are equal, for the purpose of reversing the process they must be seperate, even though knowing one means you know the other). Now, we know c), the number to find the square root of. Let's say, 64. The square root of 64 is 8. But, to figure out the equation, we have to forget that except to check to see if we have the right answer.

So, a)=? b)=? and c)=64. Is there an equation that will give a) and b) just from c)?

This equation must also work with c)=16.

Triangulation says no.

Re: OT: Interesting math problem...
 

I came here too late and missed all the fun! http://forum.shrapnelgames.com/images/smilies/wink.gif

To calculate the square root of a number, I believe it has to be done numerically, so you either have to sum an infinite series or use an iterative method that starts with an estimate and gives successively more accurate values with each iteration.

Re: OT: Interesting math problem...
 

Narf: I don't know of any cut and dry equation for finding a square root - however, there is a reasonably simple recursive method for approximating a square root to any precision desired:

R(0, N) = N
R(k, N)) = (N/R(k - 1, N) + R(k - 1, N))/2

DO NOT DO THIS RECURSIVLY - run it as a loop, saving the Last value.
Where N is the original number (constant), and k is a method to control the precision.
A sample: N = 16

Root(0, 16) = 16;
Root(1, 16) = (16/16 + 16)/2 = (1 + 16) / 2 = 8.5
Root(2, 16) = 5.1911764...
Root(3, 16) = 4.1366647...
Root(4, 16) = 4.0022575...
Root(5, 16) = 4.0000006...
Root(6, 16) = 4.0000000...

As you can see, it gets there farily quickly.

Re: OT: Interesting math problem...
 

If you're allowed to use logarithms:

sqrt (x) = e^(0.5 * ln(x))

Re: OT: Interesting math problem...
 

Thanks Jack. But, still not an equation.

Kamog, what?

Re: OT: Interesting math problem...
 

Its a system of equations, linear algebra might get it done

Re: OT: Interesting math problem...
 

Narf, the equation means you take the constant e (2.71828183...) to the power of one half times the natural logarithm of x, or log base e of x. It doesn't have to be e though, you can get away with something like 10. So, 10^(0.5 * log(x)). It works because of the laws of logarithms, since a^b=c is equivalent to log(a)(c)=b. Then, it is simple to reconstruct as a^(log(a)(c))=c. Square root is simply a number to the one half power, so 10^(0.5 * log(x)) is the sqrt(x).

Either way, you're left with pretty much imprecise methods. I remember they briefly showed how to do it in high school, and that's probably an algorithm like the one Fyron pointed to, and it would be written in an equation as the Riemann sum of varying powers of 10, which would be a very nasty looking thing. Jack's method is easier to write, so I would go with that one. To write it in terms of an equation for square root, it would be something like:

lim(x-&gt;&amp;#8734;) R(x,N) = sqrt(N)

--edit: ^ that up there is supposed to be the infinity symbol.

Re: OT: Interesting math problem...
 

Wow, math games,... What's the square root of PI?

Cheers! http://forum.shrapnelgames.com/image...ies/tongue.gif

Re: OT: Interesting math problem...
 

At my high school they taught us a way to calculate square roots. It wasn't a simple equation. More of a process that resembled long division. I can't remember it anymore though. It may have been a form of this rod thing Fyron linked to just by a different name. But it didn't look exactly like that as I recall.

Re: OT: Interesting math problem...
 

Okay, I'm an mechanical engineer and I really don't understand this sollution. How can 1023 minus 920 be 1035??????
Downright stupid if you ask me. But maybe I don't understand how it is put down. Please enlighten me.

Re: OT: Interesting math problem...
 

Ok. I guess it is some assumption grounded in years of number theory and probablility or something. http://forum.shrapnelgames.com/images/smilies/happy.gif Lacking that I went more towards the common sense, brute force method of substition. It was not inherantly obvious to me that because the trivial answer had zeros the non-trivial answer could not. I also don't see how you could make the assumption without proving or disproving posibilities that there isn't more then one correct answer.

Re: OT: Interesting math problem...
 

Timstone, you've been using computers so long you've forgotten how to do long division. http://forum.shrapnelgames.com/images/smilies/laugh.gif

1023 minus 920 is not 1035, it's 103. The 5 is brought down from the original number so you can do the next step in the process.

Re: OT: Interesting math problem...
 

Geo:
Yup, I discovered that too. But still the whole working of this sum eludes me. I can't grasp it. I must say I have only taken a quick look at it. I don't feel much for a thorough study of the phenomena. http://forum.shrapnelgames.com/images/smilies/cool.gif

Re: OT: Interesting math problem...
 

Ok, sure. I'll pretend I understand all of that, and you can have a nice, shiney medal.

/me gives Will a nice, shiney medal.

Re: OT: Interesting math problem...
 

It's also known as "lucky guess". While it is possible for a puzzle like that one to have one of the letters be zero, it is usually the case that one zero makes all zero. Plus, since we were only going for a right answer to the puzzle, I could have chose V=0 first, found it was not valid, then jumped back into V=9. But since I did V=9 first, and got a good answer, I could halt right there.

Oh, and by the way, I'm taking a pre-graduate level theoretical computer science course and a high level number theory course currently. Might have something to do with the intuition development http://forum.shrapnelgames.com/images/smilies/happy.gif

Narf:

[img]/threads/images/Graemlins/Hammer.gif[/img]

Re: OT: Interesting math problem...
 

Um...Are you hammering me, the medal, yourself or the thread? Or giving me a hammer?

Re: OT: Interesting math problem...
 

You have my sympathy, for dread fills my heart at the very thought of the hardships you are underdoing. I can but express my compassion to you, and wish you well despite your suffering.

For your heroic efforts, I hereby grant you a Medal of Gallanty. Or I would have, had Narf not already done so. Besides, we are in a bit of a shortage of medals these days.

*Ducks for cover, right behind Narf. Hammers happen to fall from the skies these days* All this message was obviously to be taken with a grain of salt, and more than a single grain if more is available.

Re: OT: Interesting math problem...
 

Kamog, I don't mean to be picky, but if you have decimal exponents available to do the e^ portion, you don't need logarithms at all:

sqrt (x) = x^(0.5)

- it's a basic identity.

Re: OT: Interesting math problem...
 

Yes.

Re: OT: Interesting math problem...
 

/me is confused.

Re: OT: Interesting math problem...
 

Quite.

Re: OT: Interesting math problem...
 

Hmm, that's a good point. http://forum.shrapnelgames.com/images/smilies/happy.gif

Yeah, I know that identity. Actually I started off with it to get the equation:
y = x^0.5
ln(y) = ln(x^0.5)
ln(y) = 0.5*ln(x)
y = e^(0.5*ln(x))

I see what you mean, why use logarithms if you have decimal exponents available.

Re: OT: Interesting math problem...
 

So, how do you calculate 64^0.5 without a calculator?

Re: OT: Interesting math problem...
 

Here's a trick to calculate square roots of square numbers without a calculator. Keep subtracting odd numbers like this:

1. 64 - 1 = 63
2. 63 - 3 = 60
3. 60 - 5 = 55
4. 55 - 7 = 48
5. 48- 9 = 39
6. 39 - 11 = 28
7. 28 - 13 = 15
8. 15 - 15 = 0

It took 8 steps to get to 0 so the square root of 64 is 8. I don't know why this works.