The following mathematical puzzle was given to one of my 7th grade tutees over the summer as an extra credit assignment. If you can solve it, I'd like to know your solution, as I could not solve it with the directions given. (The dots in the puzzle are strictly for spacing purposes, as HTML disregards excess spaces.)
There are only THREE missing digits1
. . . * * * 3
. . . x * 4 *
_____________
. . 1 * * 1 *
. * * * 9 * 0
* * * 6 * 0 0
_____________
* 4 * * 8 * *
1 These directions are suspect. I was unable to solve the puzzle under the assumption that there are only three possible replacement digits for each *. Perhaps the teacher meant that each * could be replaced by one of the digits already in the puzzle or by one of the three missing digits. However, since the only digits not represented in the puzzle already are 2, 5 and 7, there are only three missing digits anyway and the directions would be unnecessary.
4 comments:
There has to be a typo some where because...
The four times three are a given, which makes twelve. That's a 2+1 in the second column (working backwards), adding up to three, which is not one of our available choices. But it's a given in the puzzle, so it doesn't work.
Yeah, see that's why I think the directions are unclear. I think that the * can be any digits that are in the puzzle or missing from the puzzle, but there are only three of them. So, if you decide to replace one * with a 3 (even though it's already in the puzzle) you only have two other missing digits (one of which must be a 2 from the aforementioned 3x4 issue) which may or may not be in the puzzle already. Therefore, some of the other * must be 3 also. I don't know, and I couldn't get the kid's teacher to elaborate. She just said "this is a very hard puzzle that even a teacher might not be able to get". Which made me wonder why she gave it to 7th graders and also challenged me...I'm not a pushover when it comes to logic puzzles. I aced the analytical portion of the GRE.
I hadn't thought of the instructions meaning any three digits. Hmm.
Alright, looking again, it still can't work. The 4x3 is still a given, so 2 and 3 have to be two of the missing digits. In order to get the given 9 with the carried 1 from the 12, the 4 has to multiply either 2 or 7. In the product column with the 9 that adds up to something ending 8...the final result has to be 18 (even all 9's wouldn't get to 28, and 8 is obviously out of the question.) That means the other two digits in that column have to add up to the remaining 9. Using the two known missing digits (2 and 3) the final missing digit would have to be either a 6 or a 7. Neither works in the upper equation. I still say this doesn't work.
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