Can You Solve This 4th Grade Math Problem ?

Got it, but it needed some time (might be more than half an hour). It is funny to do division by hand , which you have done 30 years ago last time, thought the real challenge was the english notation I needed to look up first (in germany we use an other division notation).

The alignment is the problem for me. Spent forever trying to figure out CU, when CU is really CU0, which throws the whole alignment off

The alignment is the problem for me. Spent forever trying to figure out CU, when CU is really CU0, which throws the whole alignment off

Agreed. And I'm not used to the notation itself. It took me some time to realize that r means remainder.

This will be very useful for people who is not used to the english division representation (such as myself):

http://en.wikipedia.org/wiki/Long_division

Didn't a thread get closed a few days ago for asking us to do there homework? =-P.

But seriousky this was an fun challenge and feel a bit ashamed how long it took to actually figure out.

#Murica

Oh how the education system pales in comparison to Finland.

Thanks, Obama.

Any chance you could rewrite this in X/Y=Z notation? Following the wikipedia link makes me thing this method was created just to litter a maximum amount of text in as little space as possible, do people actually go through those intermediary steps?

It took me some time to realize that r means remainder.

Sorry, but to be honest, that's just a reading comprehension fail then.

dis homework, thread is no kill,

my threds is kill

wat?

dis homework, thread is no kill,

my threds is kill

wat?

It hasn't been locked because the moderators are actually smart enough to know that it's not the OP's homework.

dis homework, thread is no kill,

my threds is kill

wat?

It hasn't been locked because the moderators are actually smart enough to know that it's not the OP's homework.

<<< graduated high school in 1998 .

It took me some time to realize that r means remainder.

Sorry, but to be honest, that's just a reading comprehension fail then.

Maybe. In retrospect, I should've known. But in my defense, the equivalent word here which we use to describe that particular concept has more than one meaning in English, among which are "leftovers", "residue", even "fritters".

Yes, English is hard.

Hey it's a Carmen Sandiego math problem! Where has she been hiding these last 20 years!?

Maybe.

There's no maybe about it. It states quite clearly in the description of the problem that r stands for remainder. So if you didn't know that, then you failed to read it properly.

It hasn't been locked because the moderators are actually smart enough to know that it's not the OP's homework.

Shippou is actualy a forth grader, messing around, intensively smart. Breaking law

Here's how I solved it. I saw that since U is given to be 5 then S must be 2, 3, 4, 7, or 9 (can't be 1 since the dividend is a 3 digit number). Then E must be one of the remaining digits 1, 2, 3, 4, 7, 9. Just go through the combinations. Still this does take time to solve.

Took me around 12 minutes. It really came together when I realized I could do this:

A 1 2 3 4 7 9

C 1 2 3 4 7 9

E 1 2 3 4 7 9

I 1 2 3 4 7 9

N 1 2 3 4 7 9

S 1 2 3 4 7 9

U 5

And then eliminate the impossible.

By the way, Here's the problem in more sensible notation:

CAN / U = SU + E / U

CAN - CU0 = IN

IN - IU = E

Did it in under 5 minutes. The important thing is to realize that there are many clues about the relationships between digits that allow you find the solution without brute force.

[spoiler]For example, we can tell right off the bat that IU=U*U. Since we know U=5, this means IU=25 and thus I=2.

We know that IN-IU=E, which means that N=E+U, or N=E+5. The only available pairs of digits that fit this are (2,7) and (4,9). (2,7) is ruled out because 2 is I, therefore E=4 and N=9.

From CA-CU=I, we can determine than A=U+I. We already know U and I, which means A=7.

At this point all that's left are S, C, 1, and 3. We can tell that S*U=CU. Because we already know U=5, this means that S has to be odd and greater than 1. Therefore, S MUST be 3, and C=1. And voila, problem solved.
[/spoiler]

The alignment is the problem for me. Spent forever trying to figure out CU, when CU is really CU0, which throws the whole alignment off

Yes, that threw me off too. It took me almost 30 minutes overall, but the first 20 were because the alignment lead me down the wrong path. I worked through a table to eliminate digits like Dodopod, based on the relationships I could surmise from the info that was known, like Anthony suggests. I like these kinds of challenges, its like a geometric proof where you start with a couple known quantities or relationships and build to the solution.