chenxander

Oh wait, I got it.

Turn them into modular arithmetic expressions such as \(a\equiv 5\bmod{9}\)  and  \(b\equiv 33\bmod{36}\).

Then, convert them into mod 3 and mod 9 expressions in order to get the units and tens digits of the base 3 equivalents of \(a\) and \(b\).

\(a\equiv 2\bmod{3}\) --> The first digit of a in base 3 is 2.

\(a\equiv 5\bmod{9}\) --> The value of the two last digits of a in base 3 is 5. Therefore, the second digit of a in base 3 is 1.

\(b\equiv 0\bmod{3}\) --> The first digit of b in base 3 is 0.

\(b\equiv 6\bmod{9}\) --> The value of the two last digits of b in base 3 is 6. Therefore, the second digit of b in base 3 is 2.

So, we end up with 12 as the last two digits of a and 20 as the last two digits of b (both in base 3).

To subtract, we do 2 - 0 = 2 for the units digit. Since the second digit of a is less than the second digit of b, we carry over 3 from the third digit, which we must assume has a value. Then we do 4 - 2 = 2 for the second digit.

The last two digits of a - b in base 3 is 22.

Yep that's good.

If you want an answer/explanation, you need to tell us what the perimeter is.

You might have copied in LaTeX formatted numbers, so please either edit the question or provide the perimeter.

Thanks!

Thanks! It took me about five minutes of pure guessing-and-checking to get the answer but I guess it was worth it.

Hello Melody,

I understand your confusion; I believe the grid is a three-by-three grid.

The answer is 20160, but I'm not sure why.

Thanks.

Scroll down for the answer.

Since the three products are the same, it is impossible for one product to have a factor that another does not. 

Therefore, 7 and 5 don't work because they are coprime (don't share any factors) with every other number on the list.

Of course, 0 doesn't work because then one or two products will be zero while the other is a positive number.

The remaining numbers, 1, 2, 3, 4, 6, 8, and 9 , are the possible digits. 

Notice that every number only has factors of 2 and 3. So, separate the numbers into two groups depending on their factors:

factors of 2: (2,4,6,8)

factors of three: (3,6,9)

There are a total of seven factors of 2 and four factors of 3.

This is somewhat hard to explain, but this means that you must put a factor of 2 as G because otherwise one product will always have one more factor of two than another. Since G belongs to only one product, putting a factor of two as G allows the remaining six factors of 2 to be split equally between AxBxC and DxExF.

If you aren't able to completely follow along below, it's fine, as I am just guessing-and-checking. Just scroll down some more.

Try setting 2 as G. Since 8 has three factors and 2 combined with 4 has three factors, set 8 as A and D,F as 4 and 6 respectively.

You will find that this doesn't work. Instead of putting a 2 on the side, put a six as F. Then, set B as 9 and G as 3. This will balance out the factors of three evenly. 

Finally, put one as C and two as G.

My final configuration is (there are many, put they all have 2 as G):

8      3

9  2  4

1      6

Answer: The value of G is 2.

The strategy here is to find the boundary of the graph and then figure out what the actual graph is. First, plot some points for |x| + |y| = 5 on the Cartersian plane; (5,0), (4,1), (3,2), and combinations of their opposites work as well (ex. -4,1).

The graph of |x| + |y| = 5 becomes a diamond centered at the origin with perpendicular diagonals of length 10 on the x and y-axis. 

If you think about it, |x| + |y| < 5 must be all of the area within the diamond.

At this point, you can count the number of lattice points by drawing a diagram. Be careful, as the inequality sign is less than, not less than or equal to, so do not count the lattice points on the diamond itself.

By counting with symmetry, you get

6*4 + 4*4 + 1 = 41. 

The 6*4 is for the points between the axes, the 4*4 is the points on the axes (excluding the origin), and the 1 is for the origin.

 good luck