We want to calculate 259, which is just the same as 25÷9, since both of these expressions mean "how many 9s go into 25?"
The long division algorithm allows you to work this out by working from the left to the right of 25 whilst respecting place value. We normally set up the division in the following way:
9¯)25
Note the positions of the numbers!
Actually, since we want the answer to one decimal place we add as many zeroes after the decimal place to ensure we have two decimal places!
9¯)25.00
Why two? We use that extra digit to determine whether to round up or down.
The algorithm (or procedure) seems complicated at first but you might find a mnemonic helps to remember the steps. We work left to right doing the following steps
- Divide
- Multiply
- Subtract
- Bring down
and repeating until we run out of digits. The steps form the acronym DMSB. Popular mnemonics include "Does McDonalds Sell Burgers?", "Dracula Must Suck Blood" and "Dead Mice Smell Bad".
We need to know the 9 times tables or write the 9 times tables out (be repeatedly adding 9) so that we can refer to them.
1×9=92×9=183×9=274×9=365×9=456×9=547×9=638×9=729×9=81
The tens column
D: The first thing we ask ourselves is, "How many 9s go into 2?" (note this 2 actually represents 20 since it is in the tens column)
Well, none! 9 is too big to fit into 2. So we write 0 above the 2 in the tens column:
02.779¯)25.00
M: Now since 0×9=0 we write 0 underneath in the tens column:
02.779¯)25.0005.55
S: We now do the subtraction, 2−0, to determine the remainder (what remains to be divided) in the tens column.
02.779¯)25.000_5.5525.55
B: Now we bring the 5 in the ones column down next to the remainder so that it forms 25.
02.779¯)25.000_5.5525.55
The ones column
D: Now we ask ourselves, "How many 9s go into 25?" (note this 25 does actually represent 25 since it is in the ones column)
Well, 2×9=18 so 2 fit and we write 2 above the 5 in the ones column:
02.779¯)25.000_5.5525.55
M: Now since 2×9=18 we write 18 underneath in the ones column:
02.779¯)25.000_5.5525.5518.55
S: We now do the subtraction, 25−18, to determine the remainder (what remains to be divided) in the ones column.
02.779¯)25.000_5.5525.5518_.557.55
B: Now we bring the 0 in the tenths column down next to the remainder so that it forms 70.
02.779¯)25.000_5.5525.5518_.557.05
The tenths column
D: Now we ask ourselves, "How many 9s go into 70?" (note this 70 actually represents 7 since it is in the tenths column)
Well, 7×9=63 so 7 fit and we write 7 above the 0 in the tenths column:
02.779¯)25.000_5.5525.5518_.557.05
M: Now since 7×9=63 we write 63 underneath in the tenths column:
02.779¯)25.000_5.5525.5518_.557.05635
S: We now do the subtraction, 70−63, to determine the remainder (what remains to be divided) in the tenths column.
02.779¯)25.000_5.5525.5518_.557.0563_575
B: Now we bring the 0 in the hundredths column down next to the remainder so that it forms 70.
02.779¯)25.000_5.5525.5518_.557.0563_570
The hundredths column
D: Now we ask ourselves, "How many 9s go into 70?" (note this 70 actually represents 0.7 since it is in the hundredths column)
Well, 7×9=63 so 7 fit and we write 7 above the 0 in the hundredths column:
02.779¯)25.000_5.5525.5518_.557.0563_570
M: Now since 7×9=63 we write 63 underneath in the hundredths column:
02.779¯)25.000_5.5525.5518_.557.0563_57063
S: We now do the subtraction, 70−63, to determine the remainder (what remains to be divided) in the hundredths column.
02.779¯)25.000_5.5525.5518_.557.0563_57063_7
Now we could keep adding zeros and continue the procedure but we only needed to determine the second decimal place in order to correctly round to one decimal place and so we now stop the procedure.
Since the second decimal place was 7 we round up to 2.8. Therefore, 259=2.8 (1 dec. pl.).