Question

3-digit by 1-digit division with remainders - set i can use division to solve three-digit by one-digit equations with remainders.

Explanation:

The problem is about 3 - digit by 1 - digit division with remainders. Let's take an example to show how to solve such problems. Suppose we have the division problem \(345\div2\).

Step 1: Divide the first digit

We look at the first digit of the dividend, which is \(3\). We divide \(3\) by \(2\). \(2\) goes into \(3\) one time (\(2\times1 = 2\)). We write \(1\) above the \(3\) in the hundreds place. Then we subtract \(2\) from \(3\), \(3 - 2=1\).

Step 2: Bring down the next digit

We bring down the next digit of the dividend, which is \(4\), to the right of the remainder \(1\). So we now have \(14\).

Step 3: Divide the new number

We divide \(14\) by \(2\). \(2\) goes into \(14\) seven times (\(2\times7 = 14\)). We write \(7\) above the \(4\) in the tens place. Then we subtract \(14\) from \(14\), \(14 - 14 = 0\).

Step 4: Bring down the last digit

We bring down the last digit of the dividend, which is \(5\), to the right of the remainder \(0\). So we now have \(5\).

Step 5: Divide the last number

We divide \(5\) by \(2\). \(2\) goes into \(5\) two times (\(2\times2=4\)). We write \(2\) above the \(5\) in the ones place. Then we subtract \(4\) from \(5\), \(5 - 4 = 1\). This \(1\) is the remainder.

So \(345\div2 = 172\) with a remainder of \(1\).

In general, for a 3 - digit number \(abc\) (where \(a\), \(b\), \(c\) are digits and the number is \(100a + 10b + c\)) divided by a 1 - digit number \(d\) (\(d
eq0\)):

1. Divide the hundreds digit \(a\) by \(d\). Let the quotient be \(q_1\) and the remainder be \(r_1\) (\(r_1=a\bmod d\)).
2. Form a new number \(n_1=10\times r_1 + b\).
3. Divide \(n_1\) by \(d\). Let the quotient be \(q_2\) and the remainder be \(r_2\) (\(r_2 = n_1\bmod d\)).
4. Form a new number \(n_2=10\times r_2 + c\).
5. Divide \(n_2\) by \(d\). Let the quotient be \(q_3\) and the remainder be \(r_3\) (\(r_3=n_2\bmod d\)).

The final quotient is \(100q_1+10q_2 + q_3\) and the remainder is \(r_3\).

Answer:

To solve a 3 - digit by 1 - digit division with remainders, follow the long - division steps: divide the hundreds digit, bring down the tens digit, divide the new two - digit number, bring down the ones digit, and divide the new two - digit number (formed from the previous remainder and the ones digit). The result will be a quotient (a three - digit or less number) and a remainder (less than the divisor). For example, \(345\div2 = 172\) remainder \(1\).