ceiling technic

 Add the maximum possible remainder (TILE-1) before dividing!

  Simple Example with 10:

  Let's use TILE=10 for easier understanding:

  Without ceiling (wrong):

  11 / 10 = 1 (need 2!) ❌

  19 / 10 = 1 (need 2!) ❌

  With ceiling technique:

  (11 + 9) / 10 = 20 / 10 = 2 ✓

  (19 + 9) / 10 = 28 / 10 = 2 ✓

  (20 + 9) / 10 = 29 / 10 = 2 ✓ (still 2, correct!)

  (21 + 9) / 10 = 30 / 10 = 3 ✓

  Why Add (TILE-1)?

  Think of it like this:

  If remainder = 0 (perfectly divisible):

  20 / 10 = 2.0

  (20 + 9) / 10 = 29 / 10 = 2.9 → 2 (same!)

  If remainder > 0 (needs extra block):

  21 / 10 = 2.1 (remainder 1)

  (21 + 9) / 10 = 30 / 10 = 3.0 → 3 (pushed to next!)

  The Magic:

  Adding (TILE-1):

  - Remainder 0: Adds 0.9999... → stays same integer

  - Remainder ≥1: Adds enough to reach next integer

  Visual Pattern:

  Value | +9 | /10 | Result

  ------|----|----|-------

  10    | 19 | 1.9| 1 ✓

  11    | 20 | 2.0| 2 ✓ (jumps up!)

  19    | 28 | 2.8| 2 ✓

  20    | 29 | 2.9| 2 ✓

  21    | 30 | 3.0| 3 ✓ (jumps up!)

  Formula Summary:

  // Ceiling division formula:

  ceil(A/B) = (A + B - 1) / B

  // For our GEMM tiles:

  num_blocks = (matrix_size + tile_size - 1) / tile_size

  It's simple: "Add almost one tile, then divide" - this guarantees rounding up!