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When the dividend is large, estimating the quotient is not so easy. The next example shows how to
break the problem down by considering related, but smaller dividends.
Example 26. Find the quotient and remainder of the division 3060 ÷ 15.

Solution. If we read the dividend from the left, one digit at a time, we get, successively, the numbers 3,
30, 306, and 3060. Of course, according to the place-value system, these numbers stand for 3 thousands,
30 hundreds,etc.,but we do not need to think this way. To find the left-mostdigitof the quotient,
observe that the divisor, 15, does not “go into” 3, but it does go into 30 (2 times, exactly). To indicate
this we put, for the first digit of the quotient, the digit 2, directly over the right-most digit (0) in 30:

2
15 30 60

Now we compute the product 15×2 = 30 and subtract it from the initially selected part of the dividend,
i.e., from 30:

2
15 30 60
− 30
0

In this case, we get a remainder of 0. But we are not done yet. We treat this intermediate remainder
as if it were a new dividend. Does 15 “go into” 0? No – we need a larger dividend. To get one, we
“bring down” the digit 6:

2
15 3060
− 30

06

Now 06 = 6 (why?), and 15 does not go into 6. More precisely, it goesinto it 0 times. We indicate
this by putting a 0 directly above the digit 6 in the dividend:

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