Remove three 6s.
Understand the X-Wing.

Full text walkthrough

1. 0:00–0:04: start with the candidates. The video asks, “Can you remove a 6 without placing a number?” Scan columns 1 and 6 and check where a 6 can still go in these pencil marks.
2. 0:04–0:09: spot the X-Wing. Both columns have candidate 6 only in rows A and I. The four pattern cells are A1, A6, I1 and I6. The native AI Coach panel identifies the technique as X-Wing and the board highlights those cells.
3. 0:09–0:15.5: follow the reason. The two 6s must occupy opposite corners. Therefore the other cells in rows A and I cannot contain 6. The video instructs you to cross out candidate 6 in A2, A4 and I2. Keep their other candidates; this deduction does not choose which diagonal contains the 6s.

Check the two columns first

Rows A–I run from top to bottom; columns 1–9 run from left to right. In the prepared pencil marks shown:

• Column 1 can place its 6 only in A1 or I1.
• Column 6 can place its 6 only in A6 or I6.

Each base column must have exactly these two remaining positions for 6, in the same pair of rows. Four cells forming a rectangle is not sufficient by itself.

Why must the 6s take opposite corners?

Each column needs one 6, and a row cannot contain two 6s. Follow both possibilities:

1. If A1 is 6: I1 cannot be 6 because it shares column 1. Column 6 has only A6 and I6 available; row A already has its 6 at A1, so the 6 in column 6 must be I6.
2. If I1 is 6: A1 cannot be 6. Row I already has its 6, so column 6 must place its 6 at A6.

The possible pairs are A1 + I6 or I1 + A6. In either case, the 6 in each of rows A and I is already accounted for by columns 1 and 6.

Which pencil marks can go?

A2, A4 and I2 belong to those same two rows, outside the four pattern cells. A 6 in any of them would conflict with the forced pattern. Remove only their candidate 6; leave every other candidate in place.

The X-Wing itself does not tell you which diagonal is the final answer. After the elimination, look for the next justified deduction.

How were these pencil marks prepared?

A fresh candidate map still includes 6 and 9 at G1, and 6 at both C5 and G5. These two earlier deductions rebuild the complete pencil-mark state used in the video:

1. A hidden pair in the bottom-left box: within rows G–I and columns 1–3, both 3 and 5 can occur only at G1 and H1. Those two cells therefore keep only 3 and 5, removing 6 and 9 from G1.
2. A pointing pair in the centre box: within rows D–F and columns 4–6, candidate 6 occurs only at D5 and F5. Both are in column 5, so candidate 6 can be removed from both C5 and G5 outside that box.

After those steps, columns 1 and 6 each have candidate 6 only in rows A and I, establishing the X-Wing above. Use justified remaining candidates; if your candidate state differs, check the preceding deductions first.

Three checks to keep the deduction sound

• Both base columns must have exactly the same two candidate rows for 6.
• The eliminations are in cover rows A and I, outside the four pattern cells.
• Remove only 6; keep the other notes and leave the diagonal undecided.