Problem
Reasoning · Grade 3-1 Completing Equations
Ones column: a borrow is needed
5 cannot take 9, so borrowing gives 15 − 9 = 6.
This is just regrouping in subtraction: when the top digit is too small, you trade one ten for ten ones.
3.NBT.A.2Draw A DiagramWhen the top digit in the ones column is too small to subtract from, one ten is traded in for ten ones.
Why?
One ten is exactly ten ones, so moving it down changes only how the same amount is written, never how much it is.
Why?
The number is what its place amounts come to together, so shifting an amount from the tens to the ones leaves the number itself untouched.
Hundreds column tells us C, in two cases
Depending on the tens borrow, C is 2 or 1.
Whether the middle column borrows decides what is left in the hundreds, so there are just two possible values of C.
3.NBT.A.2Make A Systematic ListTens column in each case
That gives A − B = 6, or else B − A = 4.
Reading one column at a time turns the puzzle into two tidy little relationships between A and B.
3.NBT.A.2Make A Systematic ListMaximize A + B + C in each case
Maximised, the cases give 14 and 15 — the second wins.
To make a sum big you want the digits themselves big, so push A and B to their largest allowed values.
1.NBT.B.3Guess And CheckCheck the winning arrangement
Checking: 955 − 799 = 156.
Plugging the numbers back in is the surest way to know the puzzle really balances.
3.NBT.A.2Guess And CheckIf you read a subtraction one column at a time and keep track of borrowing, a 'hidden digit' puzzle is just Grade 3 subtraction you already know!
- Ones column: a borrow is needed
- Hundreds column tells us C, in two cases
- Tens column in each case
- Maximize A + B + C in each case
- Check the winning arrangement