Sum Select: rule out groups with odd and even checks.
Load a target-19 board, narrow the search to exactly one odd tile, and compare nine shortest answers with a tempting wrong total.
You do not need to add every possible group before rejecting it. An odd target must be made with an odd number of odd-valued tiles. Even tiles do not change that odd-or-even requirement. This gives you a quick filter, followed by ordinary subtraction and exact addition.
The filter is called parity. It is useful for ruling groups out, but it does not certify a correct answer. We will apply it to one fixed Sum Select board, find all nine shortest answers, and try a group that passes the filter but still misses the target.
Load a board with two odd tiles
Choose Practice this example in the side panel, or Load parity example below the Sum Select board. The target is 19. Loading replaces your current session with a single fixed practice board. Rows A–C run from top to bottom and columns 1–3 from left to right.
9 7 8 6 4 4 2 2 6
A1 and A2 are the only odd-valued tiles: 9 and 7. The other seven tiles are even. Equal values at different coordinates remain separate resources; B1 and C3 are two different 6 tiles, for example.
Count odd values, not selected tiles
Adding an even number keeps the total's parity. Adding an odd number changes it. Starting from zero, each selected odd tile switches between even and odd, so two odd tiles cancel each other's change. A sum is odd exactly when the number of selected odd tiles is odd.
This is not the same as selecting an odd number of tiles. The three tiles 8, 6, and 4 sum to 18, an even result. The two tiles 9 and 8 sum to 17, an odd result. What matters is how many selected values are odd, not the size of the selection.
For our target 19, the selected odd count must be 1, 3, 5, or another odd number. This board contains only two odd tiles, so the only possible count is exactly one. Any proposed answer that uses neither 9 nor 7, or uses both, can be rejected without adding the even tiles.
Split the search into two branches
Choose one odd tile, then find the remaining amount using only the seven even tiles:
- Include A1, the 9: exclude A2, the 7. The even tiles must supply 19 − 9 = 10.
- Include A2, the 7: exclude A1, the 9. The even tiles must supply 19 − 7 = 12.
Both remainders are even, as they must be. You may now use the usual complement method within the even pool. Do not return to the excluded odd tile to fill a remainder: that would switch the total back to even.
For a three-tile answer, choose two even tiles. The 9 branch allows 8 + 2 or 6 + 4. The 7 branch allows 8 + 4 or 6 + 6. Coordinates distinguish repeated values and prevent using one physical tile twice.
Check every shortest answer
No single tile can reach 19. The two largest tiles are 9 and 8, and their sum is only 17. Therefore no one- or two-tile selection can reach the target. The following three-tile selections do reach it, establishing three as the minimum.
| Tiles | Addition | Total |
|---|---|---|
| A1, A3, C1 | 9 + 8 + 2 | 19 |
| A1, A3, C2 | 9 + 8 + 2 | 19 |
| A1, B1, B2 | 9 + 6 + 4 | 19 |
| A1, B1, B3 | 9 + 6 + 4 | 19 |
| A1, B2, C3 | 9 + 4 + 6 | 19 |
| A1, B3, C3 | 9 + 4 + 6 | 19 |
| A2, A3, B2 | 7 + 8 + 4 | 19 |
| A2, A3, B3 | 7 + 8 + 4 | 19 |
| A2, B1, C3 | 7 + 6 + 6 | 19 |
There are six answers in the 9 branch: two physical choices for 8 + 2, and four for 6 + 4. There are three in the 7 branch: two choices for 8 + 4, and one for the two separate 6 tiles. These are all the possible even pairs for the two remainders, not just nine examples chosen from a larger set of shortest answers.
Try a wrong group that passes the filter
Select A1 and B1: 9 + 6 = 15. The group contains exactly one odd tile, so its sum is odd, just like 19. It still falls short by four. Choose Check sum: the game keeps the selection and asks you to add 4 more.
Select B2, one of the available 4 tiles, and check again. The total is now 9 + 6 + 4 = 19. The arithmetic and the parity agree, and the game accepts the answer. A parity match was necessary, but the exact total was still required.
Now choose Retry this board. It clears the selection while preserving the nine numbers and target. Try A2, B1, and C3: 7 + 6 + 6 = 19. This uses both separate 6 tiles and only one odd tile. A correct practice answer stays visible instead of advancing to a new random round.
More tiles can still form a correct answer
The game does not require the shortest selection. A2, B2, B3, C1, and C2 make 7 + 4 + 4 + 2 + 2 = 19. It is a five-tile answer, still containing exactly one odd tile. Retry and compare it with a three-tile answer if you want to check the distinction.
How many exact answers exist at all lengths?
All 512 subsets of the nine physical positions, including the empty subset, were checked. Exactly 256 contain an odd number of odd tiles; here that means one of the two odd tiles. Of those candidates, only 23 sum to 19: ten include the 9 and thirteen include the 7. Nine of those 23 use three tiles. Passing parity is therefore far weaker than having the correct total. These counts belong to this fixed board, not to every Sum Select round.
Use parity when repairing a guess
Suppose you select A1, A2, and B2: 9 + 7 + 4 = 20. This even total cannot equal the odd target. Replacing the selected 4 with an unselected 2 changes the amount by two but leaves the total even. Replacing an even tile with another even tile cannot repair this parity mismatch.
Replacing the selected 7 with an unselected 6 does change parity and gives 9 + 6 + 4 = 19. The parity observation suggests the kind of change to look for; the difference calculation and availability of an unused 6 confirm that this particular repair works.
Two self-checks
What if the paper target were 18 instead?
With an even target, the selected odd count must be even. On this board it can be zero or two. Both cases can work: A3, B1, and B2 give 8 + 6 + 4 = 18 with zero odd tiles; A1, A2, and C1 give 9 + 7 + 2 = 18 with two. This is a paper variation; the loaded parity example remains target 19.
Would three odd tiles on another board require choosing exactly one?
No. For an odd target, one or three selected odd tiles could pass the parity filter. You would still need to check their values and the exact sum. “Exactly one” was derived from the two odd tiles on this example, not a universal rule for odd targets.
Keep the filter in its proper place
On another round, identify the target's parity and count the available odd tiles before committing to a group. Then subtract, inspect the actual remaining positions, and add the proposed answer exactly. The live game reports your total but does not enforce a particular solution method or require a minimum tile count.
Every ordinary random target is constructed from three different positions, so an answer exists. The fixed example also has three-tile answers and uses the same positive 1–9 value range. Parity does not reveal which three positions generated a random target, prove that a candidate remainder is achievable, or guarantee a better result on every puzzle.
New puzzle returns to ten random rounds. Refreshing the fixed practice entry reloads this example with a clear selection; it does not save your earlier answer. This is an untimed arithmetic exercise, with no grading or health claims.