packerjg10 wrote:I was given this puzzle, and I don't even know where to begin.

Here's what you do:

Look at row 3, column 3.

Can it be a 1? No: 1 is already in row 5, column3 and you can't have two 1's in the same column.

Can it be a 2? No: 2 is at row 3, column 5 (r3c5)

Can it be a 3? No: There is already a 3 in the first box (It has 7,3,6,8,and 1)

Can it be a 4? No: r3c7

Can it be a 5? Yes, it could be.

Can it be a 6? No: r4c3

Can it be a 7? No: r1c1 (Same box)

Can it be a 8? No: r3c1 (Same box)

Can it be a 9? No: r3c4

OK, there is only one number that can go in r3c3: 5. So put a 5 there.

See? One cell solved already!

Study r4c7 the same way. You will find it must be a 7.

Another cell solved!

Now another way to find placements:

There is a 7 in r1c1 and r9c3. Now the middle box (called box 4 - has cells r4c1 and r6c3 at the corners) needs a 7. But it can't go in columns 1 or 3 as there are already 7's there. And r6c2 is already a 8. So the only cells that the 7 of box 4 can go into are r4c2 and r5c2. But row 4 already has a 7 at r4c7. So you must put the 7 in r5c2.

Another cell solved.

While looking at 7's I see that row 4 has a 7 and row 5 has a 7. The cell at r6c6 therefore must be a 7.

Good Luck. This is all you need. No "jargon" and other complexities. This puzzle can be solved as I outlined.

Mac