International Baccalaureate · PYP 5

Operations with Whole Numbers

⌚ ~66 min ▢ 6 blocks ✓ 28 exercises ➣ Prerequisites : Whole Numbers to Millions, Addition, Subtraction, Multiplication, Division

I Adding and Subtracting Large Numbers

Example

Calculate $2\,486\,357+1\,795\,648$ in columns.

Answer

Numbers in the millions are added like smaller ones: ones under ones, from right to left, with a carry to the next column whenever a column makes $10$ or more.

Ones: $7 + 8 = 15$. Write $5$ and carry $1$ ten.
Tens: $1 + 5 + 4 = 10$. Write $0$ and carry $1$ hundred.
Hundreds: $1 + 3 + 6 = 10$. Write $0$ and carry $1$ thousand.
Thousands: $1 + 6 + 5 = 12$. Write $2$ and carry $1$ ten thousand.
Ten thousands: $1 + 8 + 9 = 18$. Write $8$ and carry $1$ hundred thousand.
Hundred thousands: $1 + 4 + 7 = 12$. Write $2$ and carry $1$ million.
Millions: $1 + 2 + 1 = 4$.

So $2\,486\,357+1\,795\,648=4\,282\,005$.
Order of magnitude: $2\,500\,000+1\,800\,000=4\,300\,000$. The result $4\,282\,005$ is close to it.

Example

Calculate $6\,004\,210-2\,576\,385$ in columns.

Answer

We subtract column by column, from right to left, by compensation.

Ones: $0-5$ is not possible. Add $10$ ones to the top number, making $\textcolor{olive}{1}0$ ones, and $1$ ten to the bottom number: $\textcolor{olive}{1}0-5=5$.
Tens: $1-8-\textcolor{olive}{1}$ is not possible. Add $10$ tens to the top number, making $\textcolor{olive}{1}1$ tens, and $1$ hundred to the bottom number: $\textcolor{olive}{1}1-8-\textcolor{olive}{1}=2$.
Hundreds: $2-3-\textcolor{olive}{1}$ is not possible. Add $10$ hundreds to the top number, making $\textcolor{olive}{1}2$ hundreds, and $1$ thousand to the bottom number: $\textcolor{olive}{1}2-3-\textcolor{olive}{1}=8$.
Thousands: $4-6-\textcolor{olive}{1}$ is not possible. Add $10$ thousands to the top number, making $\textcolor{olive}{1}4$ thousands, and $1$ ten thousand to the bottom number: $\textcolor{olive}{1}4-6-\textcolor{olive}{1}=7$.
Ten thousands: $0-7-\textcolor{olive}{1}$ is not possible. Add $10$ ten thousands to the top number, making $\textcolor{olive}{1}0$ ten thousands, and $1$ hundred thousand to the bottom number: $\textcolor{olive}{1}0-7-\textcolor{olive}{1}=2$.
Hundred thousands: $0-5-\textcolor{olive}{1}$ is not possible. Add $10$ hundred thousands to the top number, making $\textcolor{olive}{1}0$ hundred thousands, and $1$ million to the bottom number: $\textcolor{olive}{1}0-5-\textcolor{olive}{1}=4$.
Millions: $6-2-\textcolor{olive}{1}=3$.

So $6\,004\,210-2\,576\,385=3\,427\,825$.
Check with an addition, the result plus the number taken away: $3\,427\,825+2\,576\,385=6\,004\,210$, the starting number.

Skills to practice

Adding and Subtracting Numbers in the Millions

II Multiplying by a Two- or Three-Digit Number

Method — Column Multiplication by a Two-Digit Number

To calculate $23 \times 37$:

Step 1: Align the numbers vertically by place value.
Step 2: Multiply by the ones digit. Multiply the top number ($\textcolor{colordef}{23}$) by the ones digit of the bottom number ($\textcolor{colordef}{7}$): $\textcolor{colordef}{23}\times\textcolor{colordef}{7}=\textcolor{colorprop}{161}$.
Step 3: Multiply by the tens digit.
- First, place a placeholder $\textcolor{olive}{0}$ in the ones column of the second row. This is because we are now multiplying by the tens digit ($\textcolor{colordef}{3}$, which represents 30). This placeholder shifts the answer one place to the left.
- Next, multiply the top number (23) by the tens digit ($\textcolor{colordef}{3}$). Calculate $\textcolor{colordef}{23}\times\textcolor{colordef}{3}=\textcolor{colorprop}{69}$ and write it to the left of the placeholder.
Step 4: Sum the partial products. Add the results from Step 2 and Step 3: $\textcolor{colordef}{161}+\textcolor{colordef}{690}=\textcolor{colorprop}{851}$
Result: $23 \times 37 = 851$.

Example

Calculate $1\,234 \times 36$ in columns.

Answer

So $1\,234\times 36=44\,424$.

Example — Multiplying by a Three-Digit Number

Compute $326\times 254$ in columns.

Answer

Multiply $326$ by each digit of $254$, one row per digit, then add the rows.

The ones digit $4$: $326\times 4=1\,304$. This is the first row.
The tens digit $5$ stands for $50$: place one placeholder $\textcolor{olive}{0}$, then write $326\times 5=1\,630$ to its left. The second row is $326\times 50=16\,300$.
The hundreds digit $2$ stands for $200$: place two placeholder zeros $\textcolor{olive}{00}$, then write $326\times 2=652$ to their left. The third row is $326\times 200=65\,200$.
Add the three partial products, aligned by place value.

$$\begin{aligned}[t] 326\times 254 &= 326\times 4+326\times 50+326\times 200 && {\scriptstyle\left(254=4+50+200\right)} \\ &= 1\,304+16\,300+65\,200 \\ &= 82\,804 \end{aligned}$$ Order of magnitude: $326$ is close to $300$ and $254$ is close to $250$, and $300\times 250=75\,000$. The result $82\,804$ is of the same order.
So $326\times 254=82\,804$.

Skills to practice

Multiplying by a Three-Digit Number
Solving Multiplication Problems

III Dividing by a Two-Digit Number

Method — The Long Division Algorithm: Multi-Steps

To divide with a remainder, like $\textcolor{olive}{130} \div \textcolor{colordef}{4}$, follow these steps:

Set up: Write the dividend ($\textcolor{olive}{130}$) inside the bracket and the divisor ($\textcolor{colordef}{4}$) outside.
Divide the first part (13): $1$ is smaller than $4$: we start with $13$ tens. “How many times does $4$ go into $13$?”$$\textcolor{colordef}{4}\times \textcolor{colorprop}{3}=\boxed{12}\ (\le \textcolor{olive}{13}),\qquad\textcolor{colordef}{4}\times \textcolor{colorprop}{4}=\cancel{16}\ (> \textcolor{olive}{13}).$$ Write $\textcolor{colorprop}{3}$ above and $12$ under $13$; then subtract.
Subtract and Bring down the next digit: $\textcolor{olive}{13}-12=\textcolor{orange}{1}$; bring down $0$ to make $\textcolor{olive}{10}$.
Divide the new number (10): “How many times does $4$ go into $10$?”$$\textcolor{colordef}{4}\times \textcolor{colorprop}{2}=\boxed{8}\ (\le \textcolor{olive}{10}),\qquad\textcolor{colordef}{4}\times \textcolor{colorprop}{3}=\cancel{12}\ (> \textcolor{olive}{10}).$$ Write $\textcolor{colorprop}{2}$ above, put $8$ under $10$, and subtract to get the remainder.
Repeat: bring down, divide, subtract, until every digit of the dividend has been used. Here no digit is left after the $0$: $\textcolor{orange}{2}$ is the remainder.
Final answer: $\textcolor{olive}{130} \div \textcolor{colordef}{4}= \divionRemainder{\textcolor{colorprop}{32}}{\textcolor{orange}{2}}$, and $\textcolor{olive}{130}=\textcolor{colordef}{4}\times\textcolor{colorprop}{32}+\textcolor{orange}{2}$.

Example

Calculate $7\,354 \div 6$.

Answer

Using long division,

we find:$$7\,354 \div 6 = \divionRemainder{1\,225}{4}.$$ Check: $6\times 1\,225=7\,350$ and $7\,350+4=7\,354$, with $4<6$.

Method — Dividing by a Two-Digit Divisor

To divide $8\,435$ by $23$:

Prepare the multiples of the divisor, once and for all: $23$, $46$, $69$, $92$, $115$, $138$, $161$, $184$, $207$.
Take the shortest starting block that is at least the divisor: $8 < 23$, so start with $84$ (with $2\,769 \div 58$, since $2 < 58$ and $27 < 58$, the starting block would be $276$). The largest multiple not exceeding $84$ is $69 = 23 \times 3$: write $3$ in the quotient and the remainder $84 - 69 = 15$.
Bring down the next digit and start again: $153$: $138 = 23 \times 6 \le 153 < 161$, write $6$, remainder $153 - 138 = 15$. Then $155$: $138 \le 155 < 161$, write $6$, remainder $155 - 138 = 17$.
Conclude and check: $8\,435 = 23 \times 366 + 17$ with $17 < 23$. Indeed $23 \times 366 = 8\,418$ and $8\,418 + 17 = 8\,435$.

If, after bringing down a digit, the number obtained is smaller than the divisor, write $0$ in the quotient and bring down the next digit: in $4\,215 \div 42$, after the first $1$, the numbers $1$ then $15$ are both smaller than $42$, so the quotient is $100$ and the remainder $15$.

Example

Divide $4\,917$ by $38$, then check.

Answer

Multiples of $38$: $38$, $76$, $114$, $152$, $190$, $228$, $266$, $304$, $342$.

$49$: $38 \le 49 < 76$, write $1$, remainder $49 - 38 = 11$.
$111$: $76 \le 111 < 114$, write $2$, remainder $111 - 76 = 35$.
$357$: $342 \le 357 < 380$, write $9$, remainder $357 - 342 = 15$.

So $4\,917 = 38 \times 129 + 15$ with $15 < 38$. Check: $38 \times 129 = 4\,902$ and $4\,902 + 15 = 4\,917$.

Skills to practice

Dividing by a Two-Digit Divisor
Solving Division Problems with a Remainder

IV Going further

Skills to practice

Solving Multi-Step Problems with Large Numbers
Quiz