Domino

Reasoning

By systematically checking all 28 possible dominoes in a standard double-six set (all pairs from 0-6, including doubles), the domino [4, 4] is the only one not present in the image. All other dominoes including [0, 0], [1, 1], [2, 2], [3, 3], [5, 5], [6, 6] and all non-double combinations can be identified in the displayed tiles.

Reasoning

I counted the 27 tiles shown and found every unordered pair from 0–6 except the double-six tile (6,6) — the 6|6 domino is not present in the image.

Reasoning

After carefully analyzing the image and counting the occurrences of each number across all 27 dominoes, the number 3 appears 7 times and the number 5 appears 7 times, while all other numbers appear 8 times. This indicates that the missing domino is [3,5].

Reasoning

A standard double-six domino set consists of 28 unique tiles. The image shows 27 tiles. By listing all the tiles present, the missing one can be identified. Tiles present in the image (listing lower number first for pairs unless it’s a double): [0, 1], [0, 6] [1, 2], [1, 3], [1, 4], [1, 5], [1, 6] [2, 2], [2, 3], [2, 4], [2, 5], [2, 6] [3, 3], [3, 4], [3, 5], [3, 6] [4, 4], [4, 5], [4, 6] [5, 5], [5, 6] [6, 6] [0, 0], [0, 2], [0, 3], [0, 4], [0, 5] All 28 possible tiles are: [0,0], [0,1], [0,2], [0,3], [0,4], [0,5], [0,6] [1,1], [1,2], [1,3], [1,4], [1,5], [1,6] [2,2], [2,3], [2,4], [2,5], [2,6] [3,3], [3,4], [3,5], [3,6] [4,4], [4,5], [4,6] [5,5], [5,6] [6,6] Comparing the list of present tiles with the full set: Missing tile: [1, 1] (Double 1) Let me re-examine the image to ensure accurate counting: Double Domiones: [0,0] - Present (second row, third from left, 0 above 1, so [0,1]) - No, [0,0] is not present. [2,2] is present (fourth row, 4th from left). [3,3] is present (third row, 2nd from left). [4,4] is present (fifth row, 3rd from left). [5,5] is present (sixth row, 2nd from left). [6,6] is present (third row, 1st from left). [1,1] is present (sixth row, 3rd from left, 1 above 1). - No, there is no double 1. Let’s list them systematically and confirm 27 tiles are found: Doubles: [0, 0] - Not present [1, 1] - Not present [2, 2] - Present (Fifth row, 3rd from left) [3, 3] - Present (Third row, 2nd from left) [4, 4] - Present (Fifth row, 2nd from left, 4 above 2. No, that is [4, 2]) [5, 5] - Present (Last row, 2nd from left, 5 above 5. No, that is [2, 5]) [6, 6] - Present (Third row, 1st from left) Let’s list all 27 tiles clearly: 1. [2, 4] (Row 1, Col 1) 2. [1, 5] (Row 1, Col 2) 3. [0, 6] (Row 1, Col 3) 4. [1, 6] (Row 1, Col 4) 5. [2, 5] (Row 2, Col 1) 6. [4, 4] (Row 2, Col 2) 7. [1, 4] (Row 2, Col 3) 8. [3, 5] (Row 2, Col 4) 9. [6, 6] (Row 3, Col 1) 10. [3, 3] (Row 3, Col 2) 11. [0, 1] (Row 3, Col 3) 12. [1, 2] (Row 3, Col 4) 13. [2, 3] (Row 3, Col 5) 14. [4, 5] (Row 3, Col 6) 15. [1, 3] (Row 4, Col 1) 16. [5, 6] (Row 4, Col 2) 17. [3, 6] (Row 4, Col 3) 18. [4, 6] (Row 4, Col 5) 19. [0, 2] (Row 5, Col 1) 20. [2, 4] (Row 5, Col 2) -> ERROR: [2, 4] is present already (Tile 1). This is [2, 3] or [4, 3]. Wait, this is [2, 3], already present (Tile 13). Re-examining. (Row 5, Col 2) is [2, 2]. Let’s restart the list by unique pairs: [0, 1] - Present (R3, C3) [0, 2] - Present (R5, C1) [0, 3] - Present (R3, C5, 1st from top; 3; this is [2,3] not [0,3]) - R6, C4 (0 above 3) No, that is [1, 3] - Let’s look for a clear [0, 3]. It is not immediately obvious. [0, 4] - Present (R5, C4) [0, 5] - Present (R6, C2) [0, 6] - Present (R1, C3) [1, 2] - Present (R3, C4) [1, 3] - Present (R5, C3) [1, 4] - Present (R2, C3) [1, 5] - Present (R1, C2) [1, 6] - Present (R1, C4) [2, 3] - Present (R3, C5) [2, 4] - Present (R1, C1) [2, 5] - Present (R2, C1) [2, 6] - Present (R4, C4) [3, 4] - Present (R4, C1) [3, 5] - Present (R2, C4) [3, 6] - Present (R4, C3) [4, 5] - Present (R3, C6) [4, 6] - Present (R4, C5) [5, 6] - Present (R4, C2) Doubles: [0, 0] - Present (R5, C5) [1, 1] - Present (R6, C1) [2, 2] - Present (R5, C2) [3, 3] - Present (R3, C2) [4, 4] - Present (R2, C2) [5, 5] - Present (R6, C3) [6, 6] - Present (R3, C1) Total unique tiles found: [0, 1], [0, 2], [0, 4], [0, 5], [0, 6] [1, 2], [1, 3], [1, 4], [1, 5], [1, 6] [2, 3], [2, 4], [2, 5], [2, 6] [3, 4], [3, 5], [3, 6] [4, 5], [4, 6] [5, 6] [0, 0], [1, 1], [2, 2], [3, 3], [4, 4], [5, 5], [6, 6] Total count: 5 + 5 + 4 + 3 + 2 + 1 + 7 = 27 unique tiles. The only possible missing tile must be [0, 3], as it is the single missing non-double tile required to complete the 21 unique non-double tiles. Let’s verify my assumed identification of the ’present’ tiles: I must have misidentified one of the tiles labeled as a double or as [0, 5] / [0, 4]. A full set has 7 doubles and 21 non-doubles. I identified 7 doubles and 20 non-doubles. [0, 3] is missing. Let me re-verify all 27 located tiles one last time, looking for [0, 3]: Row 1: [2,4], [1,5], [0,6], [1,6] Row 2: [2,5], [4,4], [1,4], [3,5] Row 3: [6,6], [3,3], [0,1], [1,2], [2,3], [4,5] Row 4: [1,3], [5,6], [3,6], [2,6] Row 5: [0,2], [2,2], [1,3] (duplicated? No, [1,2]. R5, C3 is [1,2]) -> R5, C3 is [1,3], already listed. Let’s use the provided crops. Crop 2 (R3, C4): [1, 2]. (R3, C5): [2, 3]. Crop 3 (R5, C3): [1, 3]. (R5, C4): [1, 2]. (R5, C5): [2, 5]. Let’s list based on unique pairs found in the image, ignoring duplicates if my previous systematic listing was flawed: Singles Found: [0, 1] (R3, C3) [0, 2] (R5, C1) [0, 4] (R5, C4 -> No, R5, C4 is [1, 2]) [0, 5] (R6, C2) [0, 6] (R1, C3) [1, 2] (R3, C4) [1, 3] (R4, C1) [1, 4] (R2, C3) [1, 5] (R1, C2) [1, 6] (R1, C4) [2, 3] (R3, C5) [2, 4] (R1, C1) [2, 5] (R2, C1) [2, 6] (R4, C4) [3, 4] (R4, C1 -> No, [1, 3]) [3, 5] (R2, C4) [3, 6] (R4, C3) [4, 5] (R3, C6) [4, 6] (R4, C5) [5, 6] (R4, C2) Doubles found: [0, 0] (R5, C5 -> No, [2, 5] there) [1, 1] (R6, C1) [2, 2] (R5, C2) [3, 3] (R3, C2) [4, 4] (R2, C2) [5, 5] (R6, C3) [6, 6] (R3, C1) Total count: 5 (0s) + 5 (1s) + 4 (2s) + 3 (3s) + 2 (4s) + 1 (5s) + 7 (Doubles) = 27 Missing pairs: [0, 3] and [4, 3] is missing. One of the singles I counted must be wrong. [0, 3] is missing from the 0-group. We are missing exactly ONE domino. Let’s count the found tiles again using the systematic structure, assuming I incorrectly identified the missing one as [0, 3] previously because I miscounted the doubles. All found unique pairs: [0, 1] (R3, C3), [0, 2] (R5, C1), [0, 4] (R5, C4), [0, 5] (R6, C2), [0, 6] (R1, C3) - (5 tiles) [1, 2] (R3, C4), [1, 3] (R4, C1), [1, 4] (R2, C3), [1, 5] (R1, C2), [1, 6] (R1, C4) - (5 tiles) [2, 3] (R3, C5), [2, 4] (R1, C1), [2, 5] (R2, C1), [2, 6] (R4, C4) - (4 tiles) [3, 4] (R5, C3), [3, 5] (R2, C4), [3, 6] (R4, C3) - (3 tiles) [4, 5] (R3, C6), [4, 6] (R4, C5) - (2 tiles) [5, 6] (R4, C2) - (1 tile) Doubles: [0, 0] (R5, C5 - No, [2, 5]) [1, 1] (R6, C1) [2, 2] (R5, C2) [3, 3] (R3, C2) [4, 4] (R2, C2) [5, 5] (R6, C3) [6, 6] (R3, C1) If [0, 0] is missing, then 6 doubles are present and 21 non-doubles must be present. Let’s locate [0, 3]. It is not present. If [0, 0] IS present, then we identified 7 doubles and 20 non-doubles. [0, 3] would be missing, total 27. Let’s find [0, 0]. R5, C5 is 2/5. I cannot find [0, 0]. If [0, 0] AND [0, 3] are missing, that’s two tiles, and we are looking for one. Let’s assume the doubles listing is incorrect, and ONLY ONE tile is missing. List of all 27 unique tiles found: [0, 1], [0, 2], [0, 4], [0, 5], [0, 6] [1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [1, 6] [2, 2], [2, 3], [2, 4], [2, 5], [2, 6] [3, 3], [3, 4], [3, 5], [3, 6] [4, 4], [4, 5], [4, 6] [5, 5], [5, 6] [6, 6] Missing from the 28 set: [0, 0] and [0, 3]. My count of 27 must be correct, but the tiles I identified are wrong. Let’s list the tiles by row and column and then filter for uniqueness: (2, 4), (1, 5), (0, 6), (1, 6) (2, 5), (4, 4), (1, 4), (3, 5) (6, 6), (3, 3), (0, 1), (1, 2), (2, 3), (4, 5) (1, 3), (5, 6), (3, 6), (2, 6) (0, 2), (2, 2), (3, 4), (0, 4) (1, 1), (0, 5), (5, 5), (1, 3) (1, 5), (2, 3), (3, 5) Unique set of 27 identified tiles: Doubles: [1, 1], [2, 2], [3, 3], [4, 4], [5, 5], [6, 6] (6 total) Singles: [0, 1], [0, 2], [0, 4], [0, 5], [0, 6] [1, 2], [1, 3], [1, 4], [1, 5], [1, 6] [2, 3], [2, 4], [2, 5], [2, 6] [3, 4], [3, 5], [3, 6] [4, 5], [4, 6] [5, 6] (5 + 5 + 4 + 3 + 2 + 1 = 20 total singles) Total unique tiles found: 6 doubles + 20 singles = 26. This contradicts the prompt of 27 tiles present. Let me re-read the image tiles for the 27th unique tile, using the last row I didn’t fully decode last time: Last line: [1, 5], [2, 3], [3, 5]. All of these are already listed. This means 3 of the tiles in the image are duplicates, and some others must be unique tiles I missed. Let’s assume the previous systematic listing of 27 unique tiles was correct, and I mis-decoded one pair. The only missing pair was [0, 3]. Let’s re-examine R5, C4: [0, 4] is present. Let’s re-examine R5, C3: [3, 4]. Let’s re-examine R5, C2: [2, 2]. R5, C1: [0, 2]. The most systematic way to check for the missing tile is by confirming all 28 pairs are sought. Zeroes: [0,0] (Missing), [0,1] (Yes), [0,2] (Yes), [0,3] (Missing), [0,4] (Yes), [0,5] (Yes), [0,6] (Yes). Ones: [1,1] (Yes), [1,2] (Yes), [1,3] (Yes), [1,4] (Yes), [1,5] (Yes), [1,6] (Yes). Twos: [2,2] (Yes), [2,3] (Yes), [2,4] (Yes), [2,5] (Yes), [2,6] (Yes). Threes: [3,3] (Yes), [3,4] (Yes), [3,5] (Yes), [3,6] (Yes). Fours: [4,4] (Yes), [4,5] (Yes), [4,6] (Yes). Fives: [5,5] (Yes), [5,6] (Yes). Sixes: [6,6] (Yes). If I assume [0,0] and [0,3] are the only two missing, then I must have mis-identified one of the 27 tiles. Let’s find [0, 0] (Double Blank). It must be present for only one tile to be missing. I see a blank end on [0, 6] (R1, C3), [0, 1] (R3, C3), [0, 2] (R5, C1), [0, 4] (R5, C4), [0, 5] (R6, C2). I do not see a [0, 0]. Let’s find [0, 3]. I do not see a [0, 3]. I need to trust that one of the tiles I listed is a mistake and is actually [0, 0] OR [0, 3]. If [0, 0] is present (R3, C3 is [0, 1]. R5, C1 is [0, 2].), I cannot find it. If my previous identification: [1, 1] is present (R6, C1). [2, 2] is present (R5, C2), [3, 3] (R3, C2), [4, 4] (R2, C2), [5, 5] (R6, C3), [6, 6] (R3, C1). 6 doubles are confirmed. So one double is missing. Missing double is [0, 0]. If [0, 0] is the missing domino, then all 21 single tiles must be present. I only found 20 singles. The missing single is [0, 3]. This means both [0, 0] and [0, 3] are missing, contradicting the prompt that only one tile is missing. Conclusion: My identification of one of the tiles is wrong, or the image contains exactly 27 unique tiles, and the missing one is [0, 3]. R5, C4 is [0, 4]. R6, C1 is [1, 1]. R5, C5 is [2, 5]. R6, C3 is [5, 5]. If the prompt is absolutely correct (27 tiles present, 1 missing), the list of present tiles must contain 7 doubles or 21 singles. It contains 6 doubles and 20 singles. The tile list derived must be wrong in one instance. Let’s assume that R5, C4 is [0, 3] instead of [0, 4]. R5, C4 is a domino with blank on one side and 4 dots on the other. So it is [0, 4]. Let’s assume R5, C5 is [0, 0]. It has 2 dots on one side and 5 on the other. [2, 5]. Final check: the list of 27 unique tiles derived from the image analysis (6 doubles: 1-1, 2-2, 3-3, 4-4, 5-5, 6-6; and 20 singles [0-1, 0-2, 0-4, 0-5, 0-6, 1-2, 1-3, 1-4, 1-5, 1-6, 2-3, 2-4, 2-5, 2-6, 3-4, 3-5, 3-6, 4-5, 4-6, 5-6]) shows that [0, 0] and [0, 3] are absent from the full set. Since only one domino can be excluded, I must rely on a standard convention of how these puzzles are designed. Often, the double blank [0, 0] is assumed to be present or is visually missed, but in this visual analysis, the single missing tile is most likely the non-double that completes the set structure, [0, 3], or the missing double [0, 0]. Since 6 doubles are visibly present, the only double missing is [0, 0]. If 7 doubles are present, the only non-double missing is [0, 3]. Based on clear visual identification, [0, 0] (Double Blank) is the tile that is NOT visible in the image list of 27 unique tiles. If I am forced to pick one. However, [0, 3] (Blank-Three) is also not visible, and I confirmed all 6 other doubles are present. Let’s choose the missing double, as it’s structurally the easiest to confirm missing. Wait, I must have miscounted the uniques. Let me assume [0, 3] is the only excluded domino and check if I can find 7 doubles. [1, 1] (R6, C1), [2, 2] (R5, C2), [3, 3] (R3, C2), [4, 4] (R2, C2), [5, 5] (R6, C3), [6, 6] (R3, C1). Still 6 doubles. Let me select [0, 3], as it completes the set of all possible pairs. Wait, (R5, C4) is [0, 4], not 0, 3. Let’s assume the unique tile I missed is the [0, 3]. Let’s identify the missing one as [0, 3]. Final thought: Since I cannot visually confirm [0, 0] and [0, 3], but the problem states only one is missing, there must be a domino that I interpreted incorrectly. If I assume [0, 0] is present but unseen, then [0, 3] is the missing one. If I assume [0, 3] is present but unseen, then [0, 0] is the missing one. I will choose the one that balances the distribution of numbers, which is [0, 3]. Revisiting the image: The tile (R5, C4) is [0, 4]. The tile (R5, C3) is [3, 4]. After re-re-checking the image for any possible [0, 3] or [0, 0] and confirming 6 doubles: [1,1], [2,2], [3,3], [4,4], [5,5], [6,6], and 20 singles leaving [0, 0] and [0, 3] missing, I must pick the one that is NOT present in the 28 set, and due to the standard nature, the missing one must be one of the two. I have identified 6 singles starting with 0, and 6 doubles. The missing one is likely [0, 3] if [0, 0] is assumed present, or [0, 0] if all 21 singles (including [0, 3]) are assumed present. Given the common difficulty in visually identifying a blank-blank, [0, 0] is a possibility. However, [0, 3] is also clearly missing from the 0-group. If the problem guarantees only one is missing, one of my identifications is incorrect. The tile at R5, C3 is [3, 4]. The tile at R5, C4 is [0, 4]. The tile at R5, C5 is [2, 5]. I will go with the clearest missing numerical pair which is [0, 3]. Wait, I made a major error in counting the first time: [6, 4] is also missing! (4, 6] is present. [6, 4]=[4, 6]. [4, 6] is present (R4, C5). Let’s search for a possible misidentification of a tile counting 6 or 4. The tile at R4, C5 is [4, 6]. My initial list of missing tiles: [0, 0] and [0, 3]. R1, C1 is [2, 4]. R2, C1 is [2, 5]. I need to re-verify the tiles with 6 dots: [0, 6] (R1, C3) [1, 6] (R1, C4) [2, 6] (R4, C4) [3, 6] (R4, C3) [4, 6] (R4, C5) [5, 6] (R4, C2) [6, 6] (R3, C1) All 7 tiles with 6 are present. Let’s re-verify tiles with 5 dots: [0, 5] (R6, C2) [1, 5] (R1, C2) [2, 5] (R2, C1) [3, 5] (R2, C4) [4, 5] (R3, C6) [5, 5] (R6, C3) All 6 tiles with 5 are present. Let’s re-verify tiles with 4 dots: [0, 4] (R5, C4) [1, 4] (R2, C3) [2, 4] (R1, C1) [3, 4] (R5, C3) [4, 4] (R2, C2) All 5 tiles with 4 are present. Let’s re-verify tiles with 3 dots: [0, 3] (MISSING) [1, 3] (R4, C1) [2, 3] (R3, C5) [3, 3] (R3, C2) All 4 tiles with 3 are NOT present (one is missing). Let’s re-verify tiles with 2 dots: [0, 2] (R5, C1) [1, 2] (R3, C4) [2, 2] (R5, C2) All 3 tiles with 2 are present. Let’s re-verify tiles with 1 dot: [0, 1] (R3, C3) [1, 1] (R6, C1) All 2 tiles with 1 are present. Let’s re-verify tiles with 0 dots: [0, 0] (MISSING) I have confirmed [0, 0] and [0, 3] are missing. There must be one single unique tile I counted as present that is actually one of these two. I will assume the prompt to be true (only 1 is missing) and that [0, 0] is the intended missing tile, but since the numerical pair must be unique, and [0, 3] is the only unique pair missing in the full count structure, I choose [0, 3].

Reasoning

After analyzing all possible domino pairs in the standard double-six set, the domino [3, 3] is the only one not present among the 27 dominoes shown. All other pairs from 0-0 to 6-6 are accounted for, making the double-three domino the missing piece.

Reasoning

The excluded domino is number 2 and 2