Exercises — Addition and subtraction — carrying, borrowing, word problems

Ones column: $7+8=15$. Since $15\ge 10$, a carry does happen. Split by the base into groups of ten: $15=10\cdot 1+5$, so the quotient (full tens) is $1$ and the remainder (leftover ones) is $5$. We write the remainder $r=5$ and carry the quotient $q=1$ up to the tens column. (Same split as the Division Algorithm.)

Ones column has top digit $2$ and bottom digit $9$. Since $2<9$, the top is too small, so you must borrow. After borrowing: $2+10=12$, and $12-9=3$ is written; the tens column top drops by $1$.

Align by place value (ones under ones): $\begin{array}{r}487\\ +356\end{array}$ Ones: $7+6=13=10\cdot 1+3$ → write remainder $3$, carry quotient $1$. Tens: $8+5+\underset{\text{carry}}{\underbrace{1}}=14=10\cdot 1+4$ → write $4$, carry $1$. Hundreds: $4+3+1=8<10$ → write $8$, carry $0$. Answer: $843$. Check: $843-356=487$. ✓

$\begin{array}{r}605\\ -278\end{array}$ Ones: $5-8<0$ → borrow. $5+10=15$, $15-8=7$. The tens digit now owes $1$. Tens: the $0$ owes $1$: $0-1-7<0$ → borrow again. The $0$ borrows from hundreds (becoming $10$), lends the $1$ it owed (becoming $9$), then $9-7=2$. Write $2$. Hundreds: the $6$ lent $1$ → becomes $5$. $5-2=3$. Write $3$. Answer: $327$. Check: $327+278=605$. ✓ The full borrow chain is drawn column-by-column in the figure below.

Ones: $9+1=10=10\cdot 1+0$ → write $0$, carry $1$. Tens: $9+1=10$ → write $0$, carry $1$. Hundreds: $9+1=10$ → write $0$, carry $1$. Thousands: $9+1=10$ → write $0$, carry $1$. Ten-thousands: nothing there but the carry $1$ → write $1$. Answer: $10000$. A single carry cascaded through four columns.

A column receives two digits (each $\le 9$) plus a carry-in of at most $1$. The largest total is $9+9+1=19$. Split $19$ into groups of ten: $19=10\cdot 1+9$. So the quotient (full tens) is $1$ and the remainder (what you write) is $9$. Because $19$ is less than $20$, you can never form two full tens — so the carry (the quotient) is always $0$ or $1$, never $2$ or more. This is why you only ever "carry the one." (In symbols this quotient is written $\lfloor \frac{19}{10}\rfloor$, where $\lfloor \cdot \rfloor$ just means "round down to the whole number below" — i.e. how many complete tens fit. $\lfloor \frac{19}{10}\rfloor =1$ because one full ten fits and the leftover $9$ is discarded by the rounding.)

Align: $8003-0047$. Ones: $3-7<0$ → borrow. Need $1$ from tens. Tens: tens digit is $0$ → it has nothing, so it borrows from hundreds (becoming $10$), lends $1$ down (becoming $9$). Hundreds: hundreds digit is $0$ → it too borrows from thousands (becoming $10$), lends $1$ up-neighbour (becoming $9$). Thousands: $8$ lends $1$ → becomes $7$. Now the chain stops because $7$ had something to give. Now compute: ones $13-7=6$; tens $9-4=5$; hundreds $9-0=9$; thousands $7-0=7$. Answer: $7956$. Check: $7956+47=8003$. ✓ The chain stops at the first non-zero column above the shortfall, because that column can lend without needing to borrow itself.

Three-digit numbers run $100$ to $999$. Their sum runs from $100+100=200$ up to $999+999=1998$. A 4-digit result means the sum is $\ge 1000$. The smallest 4-digit sum is $1000$ itself — reachable, e.g. $500+500=1000$. A 4-digit result requires a carry out of the hundreds column, because the thousands digit ($1$) can only appear as a carry — neither addend had a thousands digit to begin with. Answer: smallest such sum $=1000$ (from $500+500$).

Translate (see Word Problem Translation): "gave away" → subtract; "bought" → add. $1204-587+349.$ Step 1: $1204-587$ — every borrow shown. $\begin{array}{r}1204\\ -587\end{array}$

• ones: $4-7<0$ → borrow $1$ from tens: $14-7=7$. Write $7$. Tens now owes $1$.
• tens: top tens digit is $0$; it owes $1$, so $0-1-8<0$ → borrow. The $0$ borrows from hundreds (becoming $10$), lends the $1$ it owed (becoming $9$): $9-8=1$. Write $1$. Hundreds now owes $1$.
• hundreds: top hundreds digit is $2$; it owes $1$: $2-1-5<0$ → borrow. Take $10$ from thousands: $12-1-5=6$. Write $6$. Thousands now owes $1$.
• thousands: $1$ lent $1$ → becomes $0$. Write nothing (leading zero).
• So $1204-587=617$. Check: $617+587=1204$. ✓

Step 2: $617+349$.

• ones: $7+9=16=10\cdot 1+6$ → write $6$, carry $1$;
• tens: $1+4+1=6$ → write $6$, carry $0$;
• hundreds: $6+3=9$ → write $9$.
• $=966$.

Answer: $966$ books. Full check: $966+587-349=1204$. ✓

Translate: "spent" → subtract (twice); "received back / refund" → add. $1250-675-188+40.$ Step 1: $1250-675$ — every borrow shown.

• ones: $0-5<0$ → borrow $1$ from tens: $10-5=5$. Write $5$. Tens owes $1$.
• tens: top tens digit $5$ owes $1$: $5-1-7<0$ → borrow. Take $10$ from hundreds: $15-1-7=7$. Write $7$. Hundreds owes $1$.
• hundreds: top hundreds digit $2$ owes $1$: $2-1-6<0$ → borrow. Take $10$ from thousands: $12-1-6=5$. Write $5$. Thousands owes $1$.
• thousands: $1$ lent $1$ → becomes $0$. (leading zero)
• $=575$. Check: $575+675=1250$. ✓

Step 2: $575-188$ — every borrow shown.

• ones: $5-8<0$ → borrow: $15-8=7$. Write $7$. Tens owes $1$.
• tens: $7$ owes $1$: $7-1-8<0$ → borrow: $17-1-8=8$. Write $8$. Hundreds owes $1$.
• hundreds: $5$ owes $1$: $5-1-1=3$. Write $3$. No further borrow.
• $=387$. Check: $387+188=575$. ✓

Step 3: $387+40=427$ (no carry: ones $7+0=7$, tens $8+4=12$ → write $2$ carry $1$, hundreds $3+1=4$). **Answer: \427$.\ast \ast Fullcheck:$427 - 40 + 188 + 675 = 1250$. ✓

Translate: total seats available $=500\times 3=1500$ (repeated addition, see Multiplication as Repeated Addition). Tickets sold total: $312+458+269$.

• $312+458$: ones $2+8=10$ → write $0$ carry $1$; tens $1+5+1=7$; hundreds $3+4=7$ → $770$.
• $770+269$: ones $0+9=9$; tens $7+6=13$ → write $3$ carry $1$; hundreds $7+2+1=10$ → write $0$ carry $1$; thousands $1$ → $1039$.
• Sold $=1039$. Empty $=1500-1039$ — every borrow shown.
• ones: $0-9<0$ → borrow: the tens is $0$, so it chains — ones gets $10-9=1$. Write $1$.
• tens: top tens digit $0$ owes $1$: $0-1-3<0$ → borrow. Borrows from hundreds (becoming $10$), lends its owed $1$ (becoming $9$): $9-3=6$. Write $6$. Hundreds owes $1$.
• hundreds: top hundreds digit $5$ owes $1$: $5-1-0=4$. Write $4$.
• thousands: $1-0=1$. Write $1$.
• $=461$. Answer: $461$ empty seats. Check: $461+1039=1500$. ✓

Ones: $7+8=15$ → write $5$ (matches result), carry $1$. ✓ Tens: $\square +5+1(\text{carry})=?$ and the result's tens digit is $3$. So $\square +6$ must end in $3$, meaning $\square +6=13$ (we need a carry into hundreds, since $\square +6$ ranges $6$–$15$ and only $13$ ends in $3$). Thus $\square =7$, carry $1$. Hundreds check: $4+1+1(\text{carry})=6$ → matches result's $6$. ✓ Answer: $\square =7$ (the number is $477$, and $477+158=635$). ✓

Translate forward: $\text{start}-268+150=382$. Undo the operations in reverse (this is why $+/-$ are inverses): $\text{start}=382-150+268.$

• $382-150$: ones $2-0=2$; tens $8-5=3$; hundreds $3-1=2$ → $232$.
• $232+268$: ones $2+8=10$ → write $0$ carry $1$; tens $3+6+1=10$ → write $0$ carry $1$; hundreds $2+2+1=5$ → $500$. **Answer: Sam started with \500$.\ast \ast Checkforward:$500 - 268 + 150 = 382$. ✓

Write both numbers in expanded place value (from Place Value and Base-Ten System): $\overline{abc}=100a+10b+c,\overline{cba}=100c+10b+a.$ Subtract: $\overline{abc}-\overline{cba}=100a+10b+c-100c-10b-a=99a-99c=99(a-c).$ Since $a>c$, the factor $(a-c)$ is a whole number from $1$ to $9$. Multiples of $99$ from $99\cdot 1$ to $99\cdot 9$ are: $099,198,297,396,495,594,693,792,891.$ Every one has a middle digit $9$. (Reason: $99k=100k-k$; the borrow needed to compute $100k-k$ turns the tens column into a $9$ for each $k$ from $1$ to $9$.) Proved. For example $742-247=495$ — middle digit $9$. ✓