A base conversion is easy to check because it is reversible. Convert your answer back to base ten, and if you do not land on the starting number, something slipped.
This lesson belongs to SPM Mathematics number bases. It assumes you can already do conversions from base ten, and it adds the habit of proving the answer.
What is the most reliable check?
Expand the answer back to base ten. Write each digit times its place value, then add. The place values in base b are 1, b, b², b³ and so on, starting from the right.
Here is an original example. Convert 100 to base 2 using repeated division:
| Division | Quotient | Remainder |
|---|---|---|
| 100 ÷ 2 | 50 | 0 |
| 50 ÷ 2 | 25 | 0 |
| 25 ÷ 2 | 12 | 1 |
| 12 ÷ 2 | 6 | 0 |
| 6 ÷ 2 | 3 | 0 |
| 3 ÷ 2 | 1 | 1 |
| 1 ÷ 2 | 0 | 1 |
Read the remainders from the bottom up: 100 = 1100100₂.
Now expand back: 1(64) + 1(32) + 0(16) + 0(8) + 1(4) + 0(2) + 0(1) = 64 + 32 + 4 = 100. It matches, so the answer stands.
Which mistakes does the check catch?
Suppose a student reads the same remainders from the top down and writes 0010011₂. Expanding gives 16 + 2 + 1 = 19, not 100, so the order is wrong.
A different student skips the first division line, so the units digit 0 is lost, and writes 110010₂. Expanding gives 32 + 16 + 2 = 50, exactly half of 100. A result that is half or double the target points to a missing or extra digit.
A third slip uses the wrong place values, for example 1, 2, 3, 4 instead of 1, 2, 4, 8 when expanding. The check itself must use powers of the base, so write them out in a row before you start.
Which quick checks can you do before expanding?
Digit legality. Every digit must be smaller than the base. A base 5 answer containing 5, 6, 7 or 8 is wrong before you calculate anything.
Digit count. An n-digit number in base b lies between b^(n−1) and b^n − 1.
For 100 in base 2, the powers are 64 and 128, so 100 needs 7 digits. The wrong answer 110010₂ has 6 digits, so it fails at a glance.
Last digit. The units digit of the answer is the remainder when the original number is divided by the base, so 100 ÷ 2 leaves 0 and the answer must end in 0.
Worked example: find the slip in a wrong answer
A student converts 58 to base 8 and writes 27₈. Test it: 2(8) + 7(1) = 23, which is not 58.
Do the division properly. 58 ÷ 8 = 7 remainder 2, then 7 ÷ 8 = 0 remainder 7. Reading upwards, the answer is 72₈, because 7(8) + 2(1) = 58.
The student had the digits in the right set but in the wrong order. The last-digit check would also have caught it: 58 ÷ 8 leaves 2, so the answer has to end in 2.
Check yourself
A student converts 120 to base 5 and gets 404₅. Use two different checks to decide whether it is correct, then give the right answer.
Answer
Expand: 4(25) + 0(5) + 4(1) = 104, which is not 120, so the answer is wrong.
Last-digit check: 120 ÷ 5 leaves remainder 0, but 404₅ ends in 4.
Correct working: 120 ÷ 5 = 24 remainder 0, 24 ÷ 5 = 4 remainder 4, 4 ÷ 5 = 0 remainder 4. Reading upwards gives 440₅. Check: 4(25) + 4(5) + 0 = 120.
What to study next
Next, try the number bases practice set and apply an expand-back check to every answer. The mistake log tool helps you record which kind of slip you make most.
If you want a teacher to watch your working and pinpoint the error, see online one-to-one Mathematics tuition.