Problem
Reasoning · Grade 3-2 Cryptarithms and Hidden-Digit Puzzles
Use the first partial product to limit X and u
Ending in 1 needs a pair like 3 × 7.
Only a few digit pairs multiply to something ending in 1, so the very first clue already shrinks the puzzle.
3.OA.C.7Identify SubproblemsOnly a few digit pairs multiply to something ending in 1, so the very first partial product already shrinks the puzzle.
Why?
What lands in the ones place of a partial product is decided by the two ones digits alone, so that column can be judged on its own.
Why?
Sweeping every pair of digits sorts them into those that end in 1 and those that do not, with none overlooked.
Pin down X = 7 and u = 3
Staying two digits leaves only 27 × 3 = 81.
Checking the two candidates immediately rules one out because its product is too long.
3.OA.C.7Guess And CheckFind the multiplier's tens digit t
Only 27 × 6 = 162 ends in 2.
Listing the nine multiples of 27 and reading their ones digits finds the one ending in 2.
3.OA.C.7Make A Systematic ListCompute the final product and check the tens digit
So the multiplier is 63 and the product 1701.
Adding the shifted partial products is how column multiplication forms the final answer, and the tens digit 0 confirms the fit.
3.NBT.A.2Guess And CheckFill every box
1701 has 0 in the tens, matching the clue.
Every clue is satisfied at once, so the filled-in multiplication is fully determined.
3.OA.A.4Guess And CheckThe ones digit of a product is a giant clue — once you spot that 7 × 3 ends in 1, the whole hidden multiplication unlocks with Grade 3 times tables!
- Use the first partial product to limit X and u
- Pin down X = 7 and u = 3
- Find the multiplier's tens digit t
- Compute the final product and check the tens digit
- Fill every box