Multi-Digit Arithmetic

1. (a) The digit $5$ in the thousands place has a value of:

$5 \times 1{,}000 = 5{,}000$

(b) The digit $5$ in the ten-thousands place has a value of:

$5 \times 10{,}000 = 50{,}000$

(c) Compare the two values:

$\frac{50{,}000}{5{,}000} = 10$

The $5$ in the ten-thousands place is $\mathbf{10}$ times greater than the $5$ in the thousands place. This is because each place to the left is always $10$ times the place to its right.

2. (a) $612{,}904$

(b) $80{,}000 + 3{,}000 + 400 + 50 + 7 = 83{,}457$

(c) Compare the numbers:

• $53{,}192$ — five digits, starts with $53{,}1...$
• $53{,}921$ — five digits, starts with $53{,}9...$
• $503{,}921$ — six digits

From least to greatest: $53{,}192 < 53{,}921 < 503{,}921$.

3. (a) Add the pages:

$14{,}687 + 23{,}458$

Step by step:

• Ones: $7 + 8 = 15$ (write $5$, carry $1$)
• Tens: $8 + 5 + 1 = 14$ (write $4$, carry $1$)
• Hundreds: $6 + 4 + 1 = 11$ (write $1$, carry $1$)
• Thousands: $4 + 3 + 1 = 8$
• Ten-thousands: $1 + 2 = 3$

$14{,}687 + 23{,}458 = 38{,}145$

Both grades read $\mathbf{38{,}145}$ pages in total.

(b) Subtract to find the difference:

$23{,}458 - 14{,}687$

Step by step:

• Ones: $8 - 7 = 1$
• Tens: $5 - 8$ — need to regroup: $15 - 8 = 7$
• Hundreds: $3 - 6$ — need to regroup: $13 - 6 = 7$ (after borrowing)
• Thousands: $2 - 4$ — need to regroup: $12 - 4 = 8$ (after borrowing)
• Ten-thousands: $1 - 1 = 0$

$23{,}458 - 14{,}687 = 8{,}771$

The fourth graders read $\mathbf{8{,}771}$ more pages.

4. (a) Multiply $2{,}345 \times 7$:

• $5 \times 7 = 35$ (write $5$, carry $3$)
• $4 \times 7 = 28$, plus $3 = 31$ (write $1$, carry $3$)
• $3 \times 7 = 21$, plus $3 = 24$ (write $4$, carry $2$)
• $2 \times 7 = 14$, plus $2 = 16$

$2{,}345 \times 7 = 16{,}415$

The factory produces $\mathbf{16{,}415}$ toy cars in $7$ days.

(b) $16{,}415 > 15{,}000$, so yes — the warehouse will be overfull. The factory produces $16{,}415 - 15{,}000 = 1{,}415$ more cars than the warehouse can hold.

5. (a) Multiply $1{,}248 \times 4$:

• $8 \times 4 = 32$ (write $2$, carry $3$)
• $4 \times 4 = 16$, plus $3 = 19$ (write $9$, carry $1$)
• $2 \times 4 = 8$, plus $1 = 9$
• $1 \times 4 = 4$

$1{,}248 \times 4 = 4{,}992$

(b) Divide $1{,}248 \div 6$:

• $6$ goes into $12$ two times: $6 \times 2 = 12$. Remainder: $0$. Bring down $4$.
• $6$ goes into $4$ zero times. Bring down $8$.
• $6$ goes into $48$ eight times: $6 \times 8 = 48$. Remainder: $0$.

$1{,}248 \div 6 = 208$

Each branch gets $\mathbf{208}$ books.

(c) No, there are no books left over. $6 \times 208 = 1{,}248$, which is exactly the total donation.