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A random puzzle to solve...

Yeah, I've been MIA for a bit. Lots to post about, still deciding where to start. Before we actually get to a *gasp* meaningful (or at least non-fluff) post, I thought I'd throw this out there.

I came across this puzzle on Boing Boing earlier and much to my chagrin there is no solution provided so I figured I'd put it up here and see if anyone who came across it could come up with a different solution (I did email my solution to the address they have at the bottom of the story though so maybe they'll shed some light on this problem...).

Anyways, I like puzzles like these and I figured that some of you non-Boing-Boing readers might get a kick out of it as well.

Take a stab at it and then see what you think of my answer. I'm curious to know if anyone sees any glaring logical errors or stupid math mistakes (the bane of my puzzle-solving existence).

So, as I recall, you would normally solve this sort of puzzle with some simple algebra. You know that

x * y * z = 225

and you know that

x + y + z = _______

Uh, wait a minute. That won't work. Normally with 3 unknowns you can use 2 equations like that to solve them, but we actually have 4 unknowns. D'oh!

So, how else to solve the problem? Well, from my days in number sense, I recalled that 225 is actually 152 (i.e. 15 * 15). Hmmm interesting. If you break that down as far as it'll go, that works out to

3 * 5 * 3 * 5 = 225

Now you have 4 numbers that will give you the product 225. But we only want 3. So the simple solution is to combine two of the numbers and you have your ages.... except which 2 do we combine? Looks like you could do:

5 * 5 * 9 = 225 or
3 * 5 * 15 = 225 or
3 * 3 * 25 = 225

at which point you're still stuck without a definite answer. Until you think about what the puzzle actually says. A total of 3 people live at the home visited by our census taker. That implies that at least one of them would have to be an adult (i.e. over 18) and of our 3 possible solutions, only one gives us an age over 18. So

3 * 3 * 25 = 225 would tell us the ages are 3, 3, and 25

(which implies a single parent living with 2 kids which seems a bit surprising for a puzzle found in a magazine in the 1960s). So, does that follow, or did I make some really pathetic math error?


I think you're right. This is a variation on a more common puzzle, in which the "eldest" clue is needed to rule out a solution with two equal highest multipliers. In the version you found, the "eldest" clue seems unnecessary. Actually, so is the "house next door" clue, isn't it. Huh.
As one of the responses to your link points out though, 1 is a number. Leaving the possible solution of 25,9,1 in addition to 25,3,3. I don't see anything in the problem that rules that out. Am I missing something?
Damn it, I always forget the "trivial" options.

I don't see any reason why 25, 9 and 1 wouldn't be a viable option as well. There doesn't seem to be any reason it wouldn't wor, and honestly I prefer that solution (the 3 & 3 kinda bugged me in the one I came up with).


I hadn't had a chance to look at the page that caffeine_girl posted. Going by their solution, if you take into account the "eldest" question that the census taker asks, you could infer that he had to ask it because there were 2 possible options that produced the same number so:

5 + 5 + 9 = 19
3 + 5 + 15 = 23
3 + 3 + 25 = 31
1 + 9 + 25 = 35
1 + 15 + 15 = 31

Which apparently means the answer was 3, 3, and 25 after all.
Ah, nifty. I like that one better than the pop sci one.

Of course this probably demonstrates why I don't do these puzzles so much anymore. Unless I'm doing them regularly, I'm bad at remembering to account for details like the one plymouth pointed out.

I beat the smart kids! I beat the smart kids!

I beat the smart kids! I beat the smart kids!

75, 3 and 1. They're mexican and the babies live with the grandma!

Re: I beat the smart kids! I beat the smart kids!

Duh Ralph Dave, if they were mexican she'd have to be their great-great-great-grandmother. She'd only be in her mid 30s if it was their grandma.