Question

divide:
( \begin{array}{r} square \\ 303 enclose{longdiv}{834,159} end{array} ) r ( square )

Explanation:

Step1: Estimate the first digit

We divide \(834159\) by \(303\). First, look at how many times \(303\) goes into \(8341\) (since \(303\times27 = 8181\) and \(303\times28=8484\) which is too big). So the first digit of the quotient is \(27\)? Wait, no, let's do long division properly.

Wait, actually, \(303\times2753 = 303\times(2000 + 700+50 + 3)\)
\(303\times2000 = 606000\)
\(303\times700 = 212100\)
\(303\times50=15150\)
\(303\times3 = 909\)
Adding them up: \(606000+212100 = 818100\); \(818100+15150=833250\); \(833250 + 909=834159\). Wait, that's exact? Wait, let's check \(303\times2753\). Let's compute \(303\times2753\):

\(2753\times300=825900\)

\(2753\times3 = 8259\)

Then \(825900+8259 = 834159\). Oh, so there is no remainder? Wait, maybe I made a mistake in estimation. Let's do long division:

Divide \(834159\) by \(303\).

First, \(303\) into \(834\) (the first three digits of the dividend) is \(2\) times (\(303\times2 = 606\)), subtract \(606\) from \(834\) we get \(228\). Bring down the next digit \(1\) to make \(2281\).

\(303\) into \(2281\): \(303\times7 = 2121\), subtract \(2121\) from \(2281\) we get \(160\). Bring down the next digit \(5\) to make \(1605\).

\(303\) into \(1605\): \(303\times5 = 1515\), subtract \(1515\) from \(1605\) we get \(90\). Bring down the last digit \(9\) to make \(909\).

\(303\) into \(909\) is exactly \(3\) times (\(303\times3 = 909\)), subtract \(909\) from \(909\) we get \(0\).

So putting it all together, the quotient is \(2753\) and the remainder is \(0\). Wait, let's verify: \(303\times2753 = 303\times(2000 + 700 + 50 + 3)=303\times2000+303\times700+303\times50+303\times3 = 606000+212100+15150+909 = 606000+212100=818100; 818100+15150=833250; 833250+909=834159\). Yes, that's correct. So the division is exact, so the quotient is \(2753\) and the remainder is \(0\).

Wait, maybe the problem was expecting a division with remainder, but in this case, it's exact. Let's confirm the long division steps again:

1. \(303\) into \(834\) (first three digits of \(834159\)): \(303\times2 = 606\), \(834 - 606 = 228\). Bring down \(1\) to get \(2281\).

1. \(303\) into \(2281\): \(303\times7 = 2121\), \(2281 - 2121 = 160\). Bring down \(5\) to get \(1605\).

1. \(303\) into \(1605\): \(303\times5 = 1515\), \(1605 - 1515 = 90\). Bring down \(9\) to get \(909\).

1. \(303\) into \(909\): \(303\times3 = 909\), \(909 - 909 = 0\).

So the quotient is \(2753\) (from the digits we got: \(2\) (from first step), \(7\) (second), \(5\) (third), \(3\) (fourth)) and the remainder is \(0\).

Step1: Perform long division

We divide \(834159\) by \(303\):

• \(303 \times 2 = 606\), \(834 - 606 = 228\); bring down \(1\) to get \(2281\).
• \(303 \times 7 = 2121\), \(2281 - 2121 = 160\); bring down \(5\) to get \(1605\).
• \(303 \times 5 = 1515\), \(1605 - 1515 = 90\); bring down \(9\) to get \(909\).
• \(303 \times 3 = 909\), \(909 - 909 = 0\).

Combining the quotient digits (\(2, 7, 5, 3\)) gives \(2753\), and the remainder is \(0\).

Answer:

The quotient is \(2753\) and the remainder is \(0\), so the first box is \(2753\) and the second box (remainder) is \(0\).