I was writing from memory. I am sure they formulated it correctly.
I think the hardest part is framing the problem in terms of the rules of addition, once I tried to solve the problem by reducing it to:
F*10^3 + O * 10^2 + U 10 + R + (etc) = (F+F)*10^3 + (O+I)*10^2 + (etc)
It’s really messy, because while that way of representing it works, the trick is more about reasoning about the circumstances under which numbers can carry (meaning, under which circumstances (O+I)*10^2 can be expanded to (O+I - n) * 10^2 + 1 * 10^3) than it is about solving anything. There are a lot of “hidden constraints”.
First let’s get out of the way the case where letters aren’t unique, in which case the trivial case is
0000 + 0000 = 0000
or
1010 + 1010 = 2020
And similar nonsense up to F=4
Assuming they ARE unique,
R + E = E implies R is 0. This means that the carry into U+V is 0
U+V = N
O+I = I on first glance appears to imply that O = 0, and we’re stuck due to uniqueness, but that’s not necessarily true. If U+V carries, then all that is required is that (O+I) = 10+(I-1) = 9+I => O = 9.
This means that, if U+V carries, O+I carries. Which means that 2*F+1 = N
Edit: It should be noted that since O can’t be 0, at this point we know U+V must carry.
So now we have
F9U0 + FIVE = NINE
What values can F be? Well, we know that 2F+1 < 10, so F <= 4. But 8+1 = 9 = N, and O=9, since we’re maintaining uniqueness (N =/= O) we know that F <= 3.
Now we need a consistency check to narrow down F.
So we know that 2F + 1 = N and U+V-10 = N (remember that U+V must carry, hence U+V-10)
Let’s say F is 3, then N is 7. What combinations of U+V-10 = 7? The only one is 8+9 or 9+8, which is impossible, since O is 9.
So F is 2, then N is 5. Thus U+V can be 9+6 (again, impossible), or 7+8 (possible).
If F is 1, then N is 3. U+V is 9+4 (impossible), 8+5 (possible), or 7+6 (possible).
So we get:
29{7,8}0 + 2I{8,7}E = 5I5E
or
19{8,5,7,6}0 + 1I{5,8,6,7}E = 3I3E
I and E can be freely chosen from whatever remains after making your valid choice of U and V.
So F can be 1 or 2.
Ulf and Jragon - nice. So far everyone I presented it to, including myself (granted, it was a pretty small sample - maybe 10 people) just did 2 and no one got to 1.
But anyway - this was done in homework in 5th grade. I would say solving this is way more difficult than the OPs high school exam question. I know one is geometry and the other kinda algebra, but still.
They had an article on this in today’s paper here (Singapore) -
they made the point that the test is supposed to test reasoning skills more than rote regurgitation.
If true - it seems appropriate to me to accept Pi=3 as a possible solution so long as the student was explicit in setting that at the outset.
For me - I intuitively thought that the total opening area must be 1/3 of the circumference - so half of 1/3 as 1/6 I would have guessed - then thought “Nah, it can’t be that easy”
My just finished primary three (3rd grade? She’s 9)
For math they learn the “model method”. Naturally, to help he with homework I must also learn it. Must say - it is very very powerful. Using this, she can solve the types of problems I recall from age around 12 or 13 or so.
OK, the way I did the latter one:
[spoiler]was to determine that the R had to be zero so O had to be a nine so that if the U and V added up to more than 10 the second column would work. Therefore F has to be less than 4 or less. Since U=V = N+10, and the 9 is taken, the N has to be either 3 or 5, so F has to be either one or two. The answer I got was
1960
1572
3532[/spoiler]
I just wondered if there was an actual formula you could use.
The idea is to understand what you’re doing, rather than to get the right answer?
That’s what I took from the article. Rather explicitly - they said the test was to test knowledge application and reasoning ability.
To my mind, if you have got the reasoning part correct, but set Pi as 3 instead of 3.14 then so what. Putting in the “right” value of Pi is pretty damn easy and doesn’t test much of anything. Understanding how to reason and arrive at the answer is much more important - is that bad?
Is setting Pi to 3 really that much less correct than setting it to 3.1 or 3.14 or 3.141? I mean, yeah, you lose precision, but it’s a bit silly to get worked up about it. If we wanted a numerically perfect answer we’d ask for it in terms of pi. If we needed an actual calculation for real world purposes that really needed that precision, we’d probably be using a calculator or somesuch with a much more precise value.
For a problem more about conceptualization than calculation, 3 is plenty sufficient.
I wouldn’t feel like a good Doper if I didn’t pedantically point out that the first picture in the diagram is incorrect. The feature it shows as being 200 centimeters is the outside diameter, when according to the problem it should be the inside diameter.
I feel I should win at least a ham sandwich for this stunning and revelatory observation.
I think that knowing the value of pi is about more than memorization, though. It is emphatically not 3; it’s not even rational. That’s an exercise in spatial awareness and reasoning skills, too; and it’s something they should know about circles before they start trying to build revolving doors. Just two decimal places would be fine.
And I was quoting Tom Lehrer.
I get it.
You can’t take three from pi, three is less than pi, so you look at the one in the tenths place. Now, that’s really one tenth, so you change the tenth to .1, and there’s four in the hundredths place…
:smack: PI FROM THREE! PI FROM THREE!
One issue that I have with the problem as constituted is that it sets up rather arbitrary standards for what is a correct answer and doesn’t communicate it to the solvers. The moment you allow an approximation, you open a potentially thorny can of worms, to mix a couple of metaphors. Solvers should know what level of accuracy they’re expected to adhere to. I would never draw any significant conclusions from one problem, but I would be REALLY reluctant to take anything away from a problem where the guidelines aren’t made clear to the solvers.
When I first read the OP’s link, I mentally solved the problem and “gave” my answer as 200pi/6 or 100pi/3. That is, I did give it in terms of pi, because the problem asked for the maximum arc length, not a numerical approximation to the maximum arc length.
In an idealized situation, any answer other than this one is only an approximation. In a real-world situation, any answer at all is an approximation, because we can’t say that the diameter is 200 cm with infinite precision, or that the shape is a perfect circle.
It just took my senior Government class less than 5 seconds to solve it. They simplified pi to 3 to do it, but it still took less time for them to solve it than it did me to explain it.
I had the answer expressed as a product of pi in about 10 seconds. It took slightly me longer to multiply an irrational number by a repeating decimal…
The dashed line disappears into the outline of the circle. Were it to measure the outside diameter, I would expect it to look like a tangent with the dashes visible along the length of the line. That being said, I took a straight edge and held it against the line and while it covered up a portion of the outline, it did not line up with the inner circle, either. So they’re measuring neither the inner nor the outer diameter but an unspecified center diameter.
To get an idea of why this might be considered hard, watch a group of eight grade kids try to solve the problem:
You chose your user name in anticipation of this exact moment, didn’t you?
I just really want a ham sandwich.