Difficult as ABC

September 16, 2007 2:35 PM

ABC x D = BADC. What single numbers (1-9) do the letters represent? Is there any way to arrive at the answer (a) by logical deduction all the way rather than with some trial-and-error and (b) so that a 10-year-old can follow the logic?

By trial-and-error I mean, for example, having deduced that C might be, e.g., 5 (whereas it couldn't be, e.g., 3), then experimenting with various values for D (3, 7, 9) and so on. That's one way to get to an answer, but I'm hoping there's a more purely deductive route.

posted by londongeezer to Education (14 answers total)

Well, algebraically it works out too: (100*a+10*b+c) * d = 1000*b + 100*a + 10*d + c.

Now, if I'm remembering my middle school algebra right, you need three equations to calculate 4 variables. Basically you need N-1 equations to solve for N variables. I could be wrong, I only remember going over that in middle school, like I said.

In fact, an equation of that may not even be solvable at all. There may be something special about it that makes it solvable, though.
posted by delmoi at 2:43 PM on September 16, 2007

Six times any even digit yields that even digit. Five times any odd digit yields 5. That's a start.
posted by Eideteker at 2:43 PM on September 16, 2007

This question might get you on the right path. Not sure it's 10-year-old doable, but YTYOMV.
posted by wemayfreeze at 2:45 PM on September 16, 2007

Here's a really similar question that was asked recently.
posted by bluefly at 2:46 PM on September 16, 2007

delmoi: attempting to solve it algebraically misses the constraint that all the variables must be integers between zero and nine. You have to use techniques like Eideteker's to solve this sort of puzzle unless you want to get into some really advanced math.
posted by zixyer at 3:24 PM on September 16, 2007

I too brute forced it - there are no answers unless C is 0.
posted by phrontist at 3:59 PM on September 16, 2007

The answers are:
A B C D
0 0 0 0
2 1 0 6
5 1 0 3
8 6 0 8
posted by phrontist at 4:03 PM on September 16, 2007

Sideways Arithmetic from Wayside School is full of problems like this.
posted by grouse at 4:17 PM on September 16, 2007 [2 favorites]

delmoi: attempting to solve it algebraically misses the constraint that all the variables must be integers between zero and nine. You have to use techniques like Eideteker's to solve this sort of puzzle unless you want to get into some really advanced math.

Well, that's true. But I did include factors so that single-digit a,b,c and d's would still be one digit. But yeah, my solution would work with any number not just [0...9].
posted by delmoi at 4:33 PM on September 16, 2007

Response by poster: Very interesting. The instructions on my daughter's homework read explicitly: "you may use the numbers 1-9 once only". No mention of zero.

So that's one reason we found it so challenging ;)

But apart from that error by (I presume) her teacher, it does seem, possibly, a little over a ten-year-old's head. Many thanks to all.
posted by londongeezer at 4:51 PM on September 16, 2007

This may be stupid, but are we certain that ABC is 3 digits in a row and not A*B*C? Because if it's A*B*C, multiplying the whole thing by D will give the result BADC. It's just that the variables are in a new order. This is a very basic approach and I don't have the homework instructions in front of me, but I thought I'd ask.
posted by acoutu at 6:02 PM on September 16, 2007

I used to love the Wayside School books. Some of those puzzles are brutally hard.

While this particular puzzle seems to be hard because of the zero, in general I wouldn't assume a 10-year-old couldn't solve it, or something similar. Nor would I look for a step-by-step procedure to arrive at the answer. If you're solving a jigsaw you sometimes have to make trial and error guesses until the pieces fall into place. This is no different. Except your pieces are numbers and the places are A,B,C and D.

Here is how I would solve it.
ABC
x D
-------
BADC

The B at the front is probably a low number. It's certainly not 8 or 9 because there's no way D*A+carry is greater than 80. Notice the second column: D*B = D. There are lots of possibilities but B=1 is a strong one, if we assume there's no carry. To get D*C=C with no carry, and 1 already taken, we're pretty much stuck with C=0. It's the most logical guess at this point so let's try it.

Now if B is one, D*A = 10+A. (no carry from D*1=D). There's no other restriction. Let D=2. Oops, A is 10. Let D=3. A is 5.
ABCD = 5 1 0 3.

Admittedly this one is tricky (it took me a while). But this is how you would solve it; trial and error, logical guesses, process of elimination.
posted by PercussivePaul at 7:50 PM on September 16, 2007

If this was really assigned to a ten year old, I'd guess that a times sign (multiplication sign) was left out, and the actual problem meant to be assigned was ABC x D = BAD x C, or something very similar.

For a trivial solution, all digits would be 1, of course.
posted by ikkyu2 at 11:49 PM on September 16, 2007

Response by poster: acoutu - ABC is 3 digits in a row, not A*B*C
posted by londongeezer at 12:03 AM on September 17, 2007