Here is a homework problem that my neighbor’s fifth-grader was given:
ONE+FOUR=FIVE and ONE*4=FOUR
Each letter represents a digit. Solve for O.
Now, it took me, with a university degree in math&cs, about 10 minutes to solve it. Granted, I solved it for all letters, and solving for O, using a heuristic method, would probably take only a couple of minutes, but still - doesn’t it seem a bit hard for fifth grade?
I would have absolutely no clue how to go about doing that. (College algebra was my highest level, and I’ve forgotten most of the second half of it. Dimensional analysis FTW!)
But I assume (possibly in error, but one can hope) that a problem of this type was already covered in class and this is extra practice, or this is an extra credit question for smarty pants or that the teacher is hoping everyone just plays with it a bit before s/he teaches how to solve it in class on Monday, but doesn’t expect anyone to actually solve it yet.
I’ve seen all three scenarios play out with my kids’ math homework.
As far as I can tell, it’s not so much math as it is code breaking.
The easy solution:
O=0
N=0
E=0
F=0
U=0
R=0
I=0
V=0
I assume each letter stands for a unique digit, though.
Ok, you really hurt my ego here. Monday I begin my sophmore year and start pre-calc. Every college math class I’ve ever taken I’ve recieved A’s. Yet I don’t even recognise the problem above as a math problem. Do the four letters (F O U R) each represent a variable? How does one go about solving this?
Each letter represents a unique digit.
First thing you see is that F is probably 1.
Second, you can see that R has to be 0 and E has to be 5.
Then you figure that O has to be between 2 and 4 - because when multiplied by 4, ONE gives the result greater than 1000 and less than 2000.
That narrows things down. Go from there.
:smack:
Thank you. I was trying (obviously) to solve for O in the word “four”, completely overlooking the O in “one”. Had I done that I would have done exactly as you described.
Maybe I should tell my professors about this. They have been encouraging my to major in mathematics, this would burst their bubble.
I have a BA in Mathematics, and I was able to solve it, in probably 20-25 minutes.
I think that it would be great as a ‘extra credit’ problem for 5th graders, but I wouldn’t expect most of them to be able to solve it. It’s not that it requires advanced math, but it does require a bit of not-quite-conventional thinking.
For example, I got to the point where I knew that 4 * N + 2 = aU, where ‘aU’ is a 2-digit number, such that the ‘a’ would added to 4 * O, with the result being FO.
It’s solvable without making any assumptions, but it’s by no means anywhere close to straight-forward.
I did an extra credit project almost exactly like this in fifth grade math, where I created similar problems and demonstrated how to solve them. I wasn’t a genius; I got the idea from Sideways Arithmetic from Wayside School.
This is a standard sort of problem, not really a conventional algebra problem, commonly seen in crossword puzzle books (most often, consisting of a long division problem).
Google “Word arithmetic” for many hits, giving hints and clues.
Here are two:
Verbal arithmetic - Wikipedia (Wikipedia)
Solving Word Arithmetics by Jim Loy.
Furthermore, (not mentioned in exactly this form at a few sites I looked at), do this:
Draw a 10 x 10 grid. Put all the digits across the top. Put all the letters used in the problem (typically, there will be ten of them) down the side. Like this:
|0|1|2|3|4|5|6|7|8|9|
--+-+-+-+-+-+-+-+-+-+-+
A | | | | | | | | | | |
--+-+-+-+-+-+-+-+-+-+-+
B | | | | | | | | | | |
--+-+-+-+-+-+-+-+-+-+-+
C | | | | | | | | | | |
--+-+-+-+-+-+-+-+-+-+-+
D | | | | | | | | | | |
--+-+-+-+-+-+-+-+-+-+-+
E | | | | | | | | | | |
--+-+-+-+-+-+-+-+-+-+-+
F | | | | | | | | | | |
--+-+-+-+-+-+-+-+-+-+-+
G | | | | | | | | | | |
--+-+-+-+-+-+-+-+-+-+-+
H | | | | | | | | | | |
--+-+-+-+-+-+-+-+-+-+-+
J | | | | | | | | | | |
--+-+-+-+-+-+-+-+-+-+-+
K | | | | | | | | | | |
--+-+-+-+-+-+-+-+-+-+-+
Use all the hints you can figure out, to narrow down what letters could stand for what digits. If you find that a certain letter CANNOT be a certain digit, then put an X in the box for that letter/digit. (Very first hint: Multi-digit numbers are never written with a leading zero. So every letter that appears as the left-most digit of any number in the puzzle CANNOT be 0.)
If you determine, for sure, that a certain letter IS a certain digit, put a circle in that box. And since a letter can’t be two different digits, and they never have two different letters for the same digit (that’s a standard rule for these sort of puzzles), when you put a circle, you can put X in the rest of that entire row and entire column.
Continue using whatever hints you can find. (See the various web sites for lists of hints to look for.) As you gather more information, ruling in or ruling out things in your grid, the grid itself helps you find still more hints to rule in or rule out.
Continue until solved.
Ever see those logic puzzles that begin like:
“Mary, Sam, Betty, and Paul go to a movie together.
Their last names are Jones, Turner, Sampson, and Roberts (but not necessarily in that order).
Their professions are Butcher, Baker, Candle-Stick Maker, and Indian Chief (again, not necessarily in that order).
Mary stood in line behind the Baker.
Betty is married to the Indian Chief.
Mr./Mrs. Sampson had to ask Sam for change for the movie ticket . . .”
And so on. You’re supposed to figure out correctly who’s who and what their jobs are.
You do it the same way . . .
Make grids, and put X’s and O’s to indicate what’s known for sure and what’s ruled out.
For example, the statement “Mary stood in line behind the Baker” tells you that Mary isn’t the Baker.
The line Betty is married to the Indian Chief tells you that Mary isn’t the Chief. And also (assuming conventional naming conventions . . .) Wait, this tells you that they aren’t using standard naming conventions, since there’s a married couple but they all have different last names. Oh well.
You get the idea.
Word arithmetic problems are approached the same way, with the grid and the clues you can find in the digits.
Took me 10 or fifteen minutes. Here’s how I proceeded
Either R is zero, or E is zero - but if E were zero, 4 times a number that ended with zero would end in zero - so R is zero. 4*E creates a number that ends with zero, so E must be 5. F must be 1, 2 or 3 since 4 times a three digit number starts with F. Whatever F is, O must be less than 5, since O+O doesn’t lead to a carry into the thousands column. But if O is less than 5, then F must be 1, because 4 times a number less than 500 must be less than 2000. So, F is 1, and O is 2, 3 or 4. We can also see that U is even, since 4 times any number with 5 in the ones column leads to an even number in the 10s column.
Now it’s down to trying solutions, and seeing how they fit the constraints (including the constraint that no two letters are the same digit).
Let’s try O=2 and see what we get. 4*(2N5)=12U0
No matter what N is, 4 times a number less than 300 can’t equal a number greater than 1200, so O=3 or 4
If O=3, then 4*(3N5)=13U0
N could be 2, 4, 6, 7, 8, or 9
2 doesn’t work because 4325=1300 (requiring U to be zero as well as R)
If N is 4, we get 4(345)=13U0 or 1380=13U0 making U=8. So far so good.
U+N=12 so V=2 and I=O+O+the carried one = 7.
So I get 4*(345)=1380 and
1380
+345=
1725
Since this turns out to be about puzzle-solving, not math, I’m moving it from MPSIMS to the Game Room.
I had a problem similar to this as an extra credit problem in a math class.
I had to have (cough, cough) “help” from google to solve it. What can I say, my brain doesn’t work that way…
It’s actually a math problem that could be stated very formally if you had the inclination to:
For all variables, their domain is
0 <= var <= 9 (aka [0,9])
For any given string, i.e. ABCD – it can be expanded to the (A * 10[sup]3[/sup] + B * 10[sup]2[/sup] + C * 10 + D), in other words, assuming the right most digit is the zeroth digit, and as you go right it increases by one, it ends up being N * 10[sup]x[/sup] where x is the digit number.
It’s relatively simple after that, but the trick comes in realizing certain identities about addition and multiplication.
For instance:
ABC + DEF = GHI
We know, C+F = x*10 + I, where “x” is what you would “carry over”
Then
B+E+x = y*10 + H
Where x is the carried over digit, and y is what would be carried over from this computation, and it continues like this forever.
For multiplication
4BC = DEF
We know
4C = z10 + F
and
4B + z = a10 + E
etc, by the same logic.
The rest of it is just some logical intuition – i.e.
ONE + FOUR = FIVE
E + R = x * 10 + E
If the domains weren’t restricted this would be hard, for instance R=20, E=5, x=2
5+20 = 2 * 10 + 5 => 25=25 (and there are many other solutions), but if 0<=(R,E)<=9, then the only solution is x,R=0 so
E+R=x*10+E => E + 0 = 0 * 10+E => E=E.
Most of it is just intuition like that (I didn’t actually finish solving it, but I realized that it snowballs once you get the identities, the trick is remembering that you can find the values of certain variables because of the restricted domain, this problem would be way under-constrained if the domain was infinite). Though I didn’t get far enough to know whether or not it matters if you can assume the domain for O and F includes zero or not (since technically you can have a string of numbers start with 0 like 01000).
this is 5th grade, there is no need to solve the other variables, just O. students need to work out that O must be less than 5 and that the carry from Nx4 must be less than 4.
FOUR
0X4 is out, since that is too little to carry to F
1X4 is out, since that is too little to carry to F
2X4 is out, since 8 will not give an 8 again for O
3X4 is in, since 12 will give a 3 again for O if the carry from Nx4 is 1
4X4 is out, since 16 will not give a 6 for O
therefore the O=3
I actually totally missed that for F+0=F; 2O < 10
ETA: Never mind, I did get that far. It’s 2O+r<10, where r is the carry from adding N+U. What I missed is that since the carry from adding E and R =0, then r has to be either 0 or 1 – which means that the carry doesn’t matter since 24+0 and 24+1 are both under 10.
ONE+FOUR=FIVE and ONE*4=FOUR
We see that E+R equals some number that ends in R. That means R is 0.
We also see E times 4 equals some number that end in R
E=1 R=4
E=2 R=8
E=3 R=2
E=4 R=6
E=5 R=0
E=6 R=4
E=7 R=8
E=8 R=4
E=9 R=6
So E=5
N x 4 + 2 equals some number that ends U
and
N + U equals some number that ends in V
N=0 U=2 V=2 no
N=1 U=6 V=7
N=2 U=0 V=2 no
N=3 U=4 V=7
N=4 U=8 V=2
N=5 U=2 V=7 no
N=6 U=6 V=2 no
N=7 U=0 V=7 no
N=8 U=4 V=2
N=9 U=8 V=7
(I eliminated all of the sets that had duplicate numbers. I’m assuming at this point that each letter has a different value.)
N=1 U=6 V=7
N=3 U=4 V=7
N=4 U=8 V=2
N=8 U=4 V=2
N=9 U=8 V=7
O15+FO60=FI75 and O154=FO60
O35+FO40=FI75 and O354=FO40
O45+FO80=FI25 and O454=FO80
O85+FO40=FI25 and O854=FO40
O95+FO80=FI75 and O95*4=FO80
O + FO = FI so O must be less than 5
1154= 460 no
2154= 860 no
3154= 1260 no
4154= 1660 no
1354= 540 no
2354= 940 no
3354= 1340 no
4354= 1740 no
1454= 580 no
2454= 980 no
3454= 1380 yes
4454= 1780 no
1854= 740 no
2854= 1140 no
3854= 1540 no
4854= 1940 no
1954= 780 no
2954= 1180 no
3954= 1580 no
4954= 1980 no
We’re left with only one possibility that fits the pattern and doesn’t duplicate a letter/number combination. And knowing this, we can easily solve for the last remaining letter I.
So R=0, E=5, N=4, U=8, V=2, O=3, F=1, I=7
345+1380=1725 and 345*4=1380
I used to love problems like this in logic magazines when I was in elementary school.
I might have been a strange child.
Still - depending on what they’re learning in school, it isn’t necessarily that hard.
First, line them up vertically - it’s easier to see the problem:
Because E+R=E, R has to be 0, and because 4 * E ends in 0, E has to be 5.
Then, 2*O doesn’t carry - so O has to be 1-4. And since 4 x a 3 digit number that begins with 1, 2, 3, or 4 that ends up with a 4 digit number has to begin with a 1 - so F has to be 1.
From there, you can figure out that O is 3 - and you move from there.
It’s not something that is necessarily beyond a 5th grader, they know how to add and multiply. But they do need to know how to try to work through a problem from multiple directions.
It seems ridiculously hard for fifth grade. It only takes third grade arithmetic, but it takes adult logic. And the proof is that out of 18 attempts I read, only one person (shijinn) took the correct approach.
“Correct” approach? I think you may find that THAT is the only “wrong answer” possible here. The current pedagogy* in math education is that rote return of method isn’t nearly as important as having a firm internal grasp of what’s going on and how you can manipulate numbers/variables. There is no wrong approach, as long as you’re understanding the relationships between things and you eventually arrive at the correct answer. You can use manipulatives, addition, multiplication, leaps of intuitive logic…whatever works for you. Everyone who understood it got 3…nothing wrong with how they got there.
*I guess I should say a current pedagogy. That’s how my kids’ math program works. I’m sure there are others that work differently. But if the teacher mentioned in the OP gave this sort of problem with no directions, I think it’s safe to assume that’s his/her program’s philosophy, too.