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Fill in the Squares II (Posted on 2005-05-17) Difficulty: 3 of 5
Place the numbers 1 through 9 in the grid below:
   1 2 3  
A |_|_|_|  
B |_|_|_|  
C |_|_|_|  
 
  1. The product of the numbers in row C is twice the sum of the numbers in row C.
  2. The product of the numbers in column 3 is twice the sum of the numbers in column 3.
  3. If you ignore column 1, all the columns have even products.
  4. If you take the six-digit number C3 B3 B1 C1 A3 B2 and multiply it by 3, you get the six-digit number B3 B1 C1 A3 B2 C3.

Try to figure it out without using a computer program.

See The Solution Submitted by Dustin    
Rating: 2.6250 (8 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Alternate Solution | Comment 15 of 22 |

From (4), where a1=a,a2=b,a3=c,b1=d,b2=e,b3=f,c1=g,c2=h, and c3=i, 300,000i + 30,000f + 3,000d + 300g + 30c + 3e = 100,000f + 10,000d + 1,000g + 100c + 10e + i

Therefore 42,857i = 10,000f + 1,000d + 100g + 10c + e

The only values for i are 1 & 2.  i=2 doesn't meet the initial conditions, so i=1.

If i=1, that makes f=4, d=2, g=8, c=5, e=7, so filling in...

ab5

274

8h1

We know 5*4*1=20=2*(5+4+1)

Since 8*h*1=2(8+h+1) -->

8h = 18 + 2h -->

h=3, therefore filling in,

ab5

274

831

The only numbers left to fill in are 6 and 9.  b=6, because that would make column 2's product even, therefore leaving a=9.

I understand the supposed flaw in statement (3), but that statement has been explained in some of the earlier posts.

It said, "If you ignore column 1, then all the columns have even products."

This is a true statement, so, if I ignored column 1, which I did, then all the colums MUST have even products. ;)

Final answer! (It's been said before, but I gotta finish this thing)

965

274

831

That's purdy.

 


  Posted by Marc on 2005-06-03 18:23:20
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