The numbers and symbols below are placed onto six vertical strips of paper:
8 - 2 9 × 1
+ 1 = 5 1 5
= 9 0 2 ÷ 6
1 = 5 6 = 1
Rotate and rearrange the strips so that four valid equations appear across the rows.
Note: Consider the numbers as they appear on a digital watch; 0, 1, 2, 5 and 8 are the same when rotated 180°, while 6 and 9 rotate to each other.
I know this problem is about a year old, but I just wanted to show by hand that there was only one solution.
As we have all observed, the operations need to stay on the inside of the equations, so the fourth and sixth columns must be the first and last. As it turns out, no matter which way we flip them, we will always have a 9 and 1 at the ends of the top line, and a 6 and 1 at the ends of the bottom line.
The only operations that can occur in the top and bottom line are × and –. Well, let’s see what we can do.
The first line is either 9 _ _ _ _ 1 or 1 _ _ _ _ 9. Let’s run through our possibilities, starting with 9 _ _ _ _ 1.
The second slot could either be =, -, *, or another number (only 8/1 or 2/5).
If we have 9=_ _ _1, this means we either have 9=ab*1, 9=c*d1, 9=ef-1, or 9=g-h1. The only solutions would be ab=09 (no 0, so not possible), c=fraction (not possible), ef=10 (no 0, so not possible), or g=9+h1 which is a two digit number (not possible).
So 9=_ _ _ 1 is not a possible solution.
9- _ _ _1 can only be 9-_=_1 or 9-_ _=1. The only solutions to those would be 9-8=01 or 9-08=1. Either way, there is no 0 so this can’t be a solution.
9*_ _ _1 can only be 9*_=_1 or 9*_ _=1. The only solutions to those would be 9*9=81 or 9*(1/9)=1. As there is no second 9 available, the first is not a solution, and the second is of course not a solution. So this can’t be a solution.
9a_ _ _1, where a must be a number not a symbol, can only be 9a=b -1, 9a=c* 1, 9a*d=1, or 9a-e=1. The only solutions would be b=9a+1 which is a two digit number (not possible), c=9a which is a two digit number (not possible), d=fraction (not possible) or e=9a-1 which is a two digit number (not possible). So this can’t be a solution.
Therefore we can conclude that 9_ _ _ _1 is not a possible solution, not matter what you put between them.
Ok, so then we must be dealing with 1_ _ _ _9.
1=_ _ _9 can only be 1=a*b9, 1=cd*9, 1=e-f9, or 1=gh-9. The only solutions would be a=fraction (not possible), cd=fraction (not possible), e = f9+1 which is a 2 digit number (not possible), or gh=10 (no 0, so not possible). So this case is not possible.
1-_ _ _9 means we are subtracting something from 1 to get 9, or a 2 digit number that ends in 9. This is not possible.
1*_ _ _9 means we either have 1*ab=9 or 1*c=d9. The only solutions would be ab = 09 (not possible) or c=d9 which is a two digit number (not possible).
So it must be that we have 1a_ _ _9.
The only cases are 1a-b=9, 1c*d=9, 1e=f-9, or 1g=h*9. The first case can be rewritten as (10+a)-b=9 so we get b=a+1. Since we can only pick from 8/1 and 2/5, the only solution we could have is 11-2=9. The second case has d=9/1c (not possible), and the third is f=1e+9 which is a two digit number (not possible). For the fourth case, the only two digit multiple of 9 that starts with a 1 is 18. So 18=2x9 is a possible solution.
Ok, so we have 11-2=9 and 18=2x9. Let’s check out what the rest of the rows look like in those cases.
1 1 - 2 = 9
? = 1 = ÷ ?
? + 9 0 1 ?
1 8 = 5 × 6
Well, 18 does not equal 5*6, so this is not correct.
So it must be
1 8 = 2 × 9
? + 9 = 1 ?
? = 1 0 ÷ ?
1 1 - 5 = 6
Well, 11-5 is indeed 6, so we are getting somewhere.
Now we just need to place the 5 and the 6/9 in the first column and the 5 and 2 in the second column. Our only options for the second row are ?+9=12 (in which case ?=3 which isn’t possible) or ?+9=15 (in which case ?=6 which is possible).
Therefore the ONLY answer is the given solution.
1 8 = 2 × 9
6 + 9 = 1 5
5 = 1 0 ÷ 2
1 1 - 5 = 6
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Posted by nikki
on 2005-01-07 19:41:56 |