Find the missing piece of a sum.
Use remainders, check every two-tile answer on a worked board, and distinguish equal values from reusable tiles.
When nine numbers are available, there are many possible groups. You can make the search smaller by turning the target into a remainder. Choose a number, subtract it from the target, and ask a simpler question: which tiles can make what is left?
Begin with one useful subtraction
Suppose the target is 17 and the board contains these tiles:
9 4 7 2 8 3 6 1 5
Start with 9. The remainder is 8, and an 8 is visible, so 9 + 8 solves the round in two tiles. You do not have to use three numbers just because the game generates its targets from three starting tiles. Any nonempty selection with the exact total is accepted.
When the remainder is not a single tile
Choose 7 instead. The remainder is 10. No tile has a value of 10, but pairs such as 9 + 1, 8 + 2, and 6 + 4 all make it. Together with the selected 7, each pair produces a valid solution. Keep track of physical tiles: you cannot reuse the same tile twice.
Now choose 8 and 6. Their total is 14, so the remainder is 3. Selecting the 3 finishes the target. Working from a partial selection often makes the next choice obvious.
Repair a total that is too large
Suppose you have selected 9, 7, and 4 for a total of 20. The target is 17, so your group is three too large. Removing a selected 3 would help, but no 3 is currently selected. Instead, replace the selected 7 with the unselected 4 only if there is a second 4 available. On this particular board there is not, because the only 4 is already selected.
A valid adjustment here is replacing the selected 4 with the unselected 1. The new group is 9 + 7 + 1 = 17. This example shows why differences and tile availability need to be checked together.
Work through a board with repeated values
Here is a constructed teaching board with a target of 10. Choose Practice this example beside this guide, or Load remainder example on the Sum Select game page, to load these exact numbers. Loading replaces the current game session. You can also use a paper copy. Name the rows A, B, and C from top to bottom, and the columns 1, 2, and 3 from left to right. A1 is the upper-left tile; A2 is the tile beside it.
5 5 8 2 7 3 6 4 1
No single tile shows 10. For a two-tile answer, the second value must equal 10 minus the first. Work across the board in order, and only look at positions later than the tile you are checking. That prevents counting A1 with A2 and then counting A2 with A1 as a different answer. It also prevents using a tile as its own partner.
- A1 with A2: 10 − 5 = 5, and a second 5 is available. The two different tiles make 5 + 5 = 10.
- A3 with B1: 10 − 8 = 2. Those tiles make 8 + 2 = 10.
- B2 with B3: 10 − 7 = 3. Those tiles make 7 + 3 = 10.
- C1 with C2: 10 − 6 = 4. Those tiles make 6 + 4 = 10.
For the other starting positions, there is no new partner later in the scan. A2 has no further 5; B1 has no later 8; B3 has no later 7; C2 has no later 6. C3 would need a 9, which is absent. These are therefore all four two-tile selections for this exact board, and two tiles is the shortest possible answer because no one-tile answer exists.
A longer answer still works. B1, B2, and C3 give 2 + 7 + 1 = 10. The game accepts an exact nonempty total; it does not demand the fewest tiles. This exercise distinguishes finding an answer from establishing that an answer uses the minimum number of tiles.
Equal values are separate resources
A1 and A2 have the same value but different addresses. Selecting both is allowed. Selecting A1 twice cannot contribute 5 twice: in the game, selecting a chosen tile again removes it from the group. If only one 5 were available, 5 + 5 would not be a usable two-tile answer, even though the arithmetic is correct.
If the 5 at A2 were unavailable, would this target become impossible?
No. The 8 + 2, 7 + 3, and 6 + 4 pairs use other tiles, so each still reaches 10. The three-tile answer 2 + 7 + 1 also remains available. Losing one proposed answer is not the same as losing every answer; check the remaining physical tiles.
Try a three-stage practice routine
- Look for a single tile equal to the target.
- Try complements: for each tile, check whether target minus that tile is available elsewhere.
- If no pair is obvious, choose two tiles and calculate the remaining amount.
This routine is a search strategy, not a guarantee that the shortest answer is found first. Longer combinations may be valid too. In our game, every target has at least one three-tile solution, so you can always return to examining triples when a simpler answer is not apparent.
Compare answers on the same board
In the fixed practice board, select A1 and A2 to make 5 + 5, then choose Check sum. After the correct answer remains visible, choose Retry this board. It clears your selection while keeping the same nine numbers and target. Try A3 and B1 for 8 + 2, then compare the other two pairs above. Practice stays on one board; it does not advance through the ten random rounds. New puzzle returns to the normal random session.
Keep the round untimed
The running total is there to support your reasoning. Use it to check a prediction, then adjust a mistaken selection. Incorrect checks do not deduct points or remove tiles. The useful question is how you found the total, not how quickly you can press the button.