Question

1. \\(2\enclose{longdiv}{419}\\) 4. \\(9\enclose{longdiv}{217}\\) 7. \\(3\enclose{longdiv}{6334}\\) 8. \\(5\enclose{longdiv}{8248}\\) 11. \\(7\enclose{longdiv}{4999}\\) 12. \\(6\enclose{longdiv}{4272}\\) 15. \\(5\enclose{longdiv}{4271}\\) 16. \\(4\enclose{longdiv}{1847}\\)

Explanation:

Let's solve problem 3: \(2\enclose{longdiv}{419}\) (assuming it's \(419\div2\))

Step1: Divide 4 by 2

\(4\div2 = 2\), write 2 above the 4.

Step2: Multiply 2 by 2

\(2\times2 = 4\), subtract from 4: \(4 - 4 = 0\).

Step3: Bring down 1

Now we have 1. \(1\div2\) is 0 (since 2 > 1), write 0 above the 1.

Step4: Bring down 9

Now we have 19. \(19\div2 = 9\) with a remainder of 1 (since \(2\times9 = 18\), \(19 - 18 = 1\)).
So \(419\div2 = 209\) with a remainder of 1, or as a decimal \(209.5\).

(Note: For other problems, follow similar long division steps. For example, problem 7: \(3\enclose{longdiv}{6334}\))

Step1: Divide 6 by 3

\(6\div3 = 2\), write 2 above the 6.

Step2: Multiply 2 by 3

\(2\times3 = 6\), subtract: \(6 - 6 = 0\).

Step3: Bring down 3

\(3\div3 = 1\), write 1 above the 3.

Step4: Multiply 1 by 3

\(1\times3 = 3\), subtract: \(3 - 3 = 0\).

Step5: Bring down 3

\(3\div3 = 1\), write 1 above the 3.

Step6: Multiply 1 by 3

\(1\times3 = 3\), subtract: \(3 - 3 = 0\).

Step7: Bring down 4

\(4\div3 = 1\) with a remainder of 1 (since \(3\times1 = 3\), \(4 - 3 = 1\)).
So \(6334\div3 = 2111\) with a remainder of 1, or \(2111.\overline{3}\).

(We can solve each problem similarly using long division steps for division of integers or decimals.)

Answer:

For problem 3 (\(419\div2\)): \(209\) remainder \(1\) or \(209.5\)
For problem 7 (\(6334\div3\)): \(2111\) remainder \(1\) or \(2111.\overline{3}\)
(And so on for other problems following the same long - division method)