GCSE Maths: Recurring decimals and fractions (Higher)

Method. Simplify the fraction, write its denominator as a product of primes, and see whether any prime other than 2 or 5 appears.

1. Say what the test is. A fraction in its simplest form gives a terminating decimal only if its denominator has no prime factors other than 2 and 5. This is because 10 = 2 × 5, so only denominators made of 2s and 5s can be turned into tenths, hundredths and so on.
   38
2. Check the fraction is in its simplest form. Common factors could change the denominator, so we check first.
   3 and 8 share no factor except 1, so 38 is already simplest.
3. Break the denominator into prime factors. A prime factor is a prime number that divides into it. We divide by the smallest prime each time.
   8 = 2³
   8 ÷ 2 = 4
   4 ÷ 2 = 2
   2 ÷ 2 = 1
4. Look for primes other than 2 and 5. If there is any other prime factor, the decimal can never end.
   Only 2s and 5s appear, so the decimal terminates.
5. Write the conclusion. Answer the question that was asked.
   38 gives a terminating decimal.

Check. A different route: divide. 3 ÷ 8 = 0.375 (it ends). ✓

Watch out. Always simplify first. For example 6/15 = 2/5 terminates even though 15 has a factor of 3.

Method. Simplify the fraction, write its denominator as a product of primes, and see whether any prime other than 2 or 5 appears.

1. Say what the test is. A fraction in its simplest form gives a terminating decimal only if its denominator has no prime factors other than 2 and 5. This is because 10 = 2 × 5, so only denominators made of 2s and 5s can be turned into tenths, hundredths and so on.
   712
2. Check the fraction is in its simplest form. Common factors could change the denominator, so we check first.
   7 and 12 share no factor except 1, so 712 is already simplest.
3. Break the denominator into prime factors. A prime factor is a prime number that divides into it. We divide by the smallest prime each time.
   12 = 2² × 3
   12 ÷ 2 = 6
   6 ÷ 2 = 3
   3 ÷ 3 = 1
4. Look for primes other than 2 and 5. If there is any other prime factor, the decimal can never end.
   The prime 3 appears, so the decimal recurs.
5. Write the conclusion. Answer the question that was asked.
   712 gives a recurring decimal.

Check. A different route: divide. 7 ÷ 12 = 0.583… (a block of digits repeats). ✓

Watch out. Always simplify first. For example 6/15 = 2/5 terminates even though 15 has a factor of 3.

Method. Simplify the fraction, write its denominator as a product of primes, and see whether any prime other than 2 or 5 appears.

1. Say what the test is. A fraction in its simplest form gives a terminating decimal only if its denominator has no prime factors other than 2 and 5. This is because 10 = 2 × 5, so only denominators made of 2s and 5s can be turned into tenths, hundredths and so on.
   920
2. Check the fraction is in its simplest form. Common factors could change the denominator, so we check first.
   9 and 20 share no factor except 1, so 920 is already simplest.
3. Break the denominator into prime factors. A prime factor is a prime number that divides into it. We divide by the smallest prime each time.
   20 = 2² × 5
   20 ÷ 2 = 10
   10 ÷ 2 = 5
   5 ÷ 5 = 1
4. Look for primes other than 2 and 5. If there is any other prime factor, the decimal can never end.
   Only 2s and 5s appear, so the decimal terminates.
5. Write the conclusion. Answer the question that was asked.
   920 gives a terminating decimal.

Check. A different route: divide. 9 ÷ 20 = 0.45 (it ends). ✓

Watch out. Always simplify first. For example 6/15 = 2/5 terminates even though 15 has a factor of 3.

Method. Simplify the fraction, write its denominator as a product of primes, and see whether any prime other than 2 or 5 appears.

1. Say what the test is. A fraction in its simplest form gives a terminating decimal only if its denominator has no prime factors other than 2 and 5. This is because 10 = 2 × 5, so only denominators made of 2s and 5s can be turned into tenths, hundredths and so on.
   511
2. Check the fraction is in its simplest form. Common factors could change the denominator, so we check first.
   5 and 11 share no factor except 1, so 511 is already simplest.
3. Break the denominator into prime factors. A prime factor is a prime number that divides into it. We divide by the smallest prime each time.
   11 = 11
   11 ÷ 11 = 1
4. Look for primes other than 2 and 5. If there is any other prime factor, the decimal can never end.
   The prime 11 appears, so the decimal recurs.
5. Write the conclusion. Answer the question that was asked.
   511 gives a recurring decimal.

Check. A different route: divide. 5 ÷ 11 = 0.45… (a block of digits repeats). ✓

Watch out. Always simplify first. For example 6/15 = 2/5 terminates even though 15 has a factor of 3.

Method. Simplify the fraction, write its denominator as a product of primes, and see whether any prime other than 2 or 5 appears.

1. Say what the test is. A fraction in its simplest form gives a terminating decimal only if its denominator has no prime factors other than 2 and 5. This is because 10 = 2 × 5, so only denominators made of 2s and 5s can be turned into tenths, hundredths and so on.
   1340
2. Check the fraction is in its simplest form. Common factors could change the denominator, so we check first.
   13 and 40 share no factor except 1, so 1340 is already simplest.
3. Break the denominator into prime factors. A prime factor is a prime number that divides into it. We divide by the smallest prime each time.
   40 = 2³ × 5
   40 ÷ 2 = 20
   20 ÷ 2 = 10
   10 ÷ 2 = 5
   5 ÷ 5 = 1
4. Look for primes other than 2 and 5. If there is any other prime factor, the decimal can never end.
   Only 2s and 5s appear, so the decimal terminates.
5. Write the conclusion. Answer the question that was asked.
   1340 gives a terminating decimal.

Check. A different route: divide. 13 ÷ 40 = 0.325 (it ends). ✓

Watch out. Always simplify first. For example 6/15 = 2/5 terminates even though 15 has a factor of 3.

Method. Simplify the fraction, write its denominator as a product of primes, and see whether any prime other than 2 or 5 appears.

1. Say what the test is. A fraction in its simplest form gives a terminating decimal only if its denominator has no prime factors other than 2 and 5. This is because 10 = 2 × 5, so only denominators made of 2s and 5s can be turned into tenths, hundredths and so on.
   415
2. Check the fraction is in its simplest form. Common factors could change the denominator, so we check first.
   4 and 15 share no factor except 1, so 415 is already simplest.
3. Break the denominator into prime factors. A prime factor is a prime number that divides into it. We divide by the smallest prime each time.
   15 = 3 × 5
   15 ÷ 3 = 5
   5 ÷ 5 = 1
4. Look for primes other than 2 and 5. If there is any other prime factor, the decimal can never end.
   The prime 3 appears, so the decimal recurs.
5. Write the conclusion. Answer the question that was asked.
   415 gives a recurring decimal.

Check. A different route: divide. 4 ÷ 15 = 0.26… (a block of digits repeats). ✓

Watch out. Always simplify first. For example 6/15 = 2/5 terminates even though 15 has a factor of 3.

Method. Simplify the fraction, write its denominator as a product of primes, and see whether any prime other than 2 or 5 appears.

1. Say what the test is. A fraction in its simplest form gives a terminating decimal only if its denominator has no prime factors other than 2 and 5. This is because 10 = 2 × 5, so only denominators made of 2s and 5s can be turned into tenths, hundredths and so on.
   1725
2. Check the fraction is in its simplest form. Common factors could change the denominator, so we check first.
   17 and 25 share no factor except 1, so 1725 is already simplest.
3. Break the denominator into prime factors. A prime factor is a prime number that divides into it. We divide by the smallest prime each time.
   25 = 5²
   25 ÷ 5 = 5
   5 ÷ 5 = 1
4. Look for primes other than 2 and 5. If there is any other prime factor, the decimal can never end.
   Only 2s and 5s appear, so the decimal terminates.
5. Write the conclusion. Answer the question that was asked.
   1725 gives a terminating decimal.

Check. A different route: divide. 17 ÷ 25 = 0.68 (it ends). ✓

Watch out. Always simplify first. For example 6/15 = 2/5 terminates even though 15 has a factor of 3.

Method. Simplify the fraction, write its denominator as a product of primes, and see whether any prime other than 2 or 5 appears.

1. Say what the test is. A fraction in its simplest form gives a terminating decimal only if its denominator has no prime factors other than 2 and 5. This is because 10 = 2 × 5, so only denominators made of 2s and 5s can be turned into tenths, hundredths and so on.
   56
2. Check the fraction is in its simplest form. Common factors could change the denominator, so we check first.
   5 and 6 share no factor except 1, so 56 is already simplest.
3. Break the denominator into prime factors. A prime factor is a prime number that divides into it. We divide by the smallest prime each time.
   6 = 2 × 3
   6 ÷ 2 = 3
   3 ÷ 3 = 1
4. Look for primes other than 2 and 5. If there is any other prime factor, the decimal can never end.
   The prime 3 appears, so the decimal recurs.
5. Write the conclusion. Answer the question that was asked.
   56 gives a recurring decimal.

Check. A different route: divide. 5 ÷ 6 = 0.83… (a block of digits repeats). ✓

Watch out. Always simplify first. For example 6/15 = 2/5 terminates even though 15 has a factor of 3.

Method. A fraction whose denominator has a prime factor other than 2 or 5 gives a recurring decimal. Use long division and watch for a remainder that repeats.

1. Say what the fraction means. The fraction bar means divide. The denominator has a prime factor other than 2 or 5, so we expect the decimal to repeat.
   211 = 2 ÷ 11
2. Set up the long division 2 ÷ 11. Write the number with a decimal point and extra zeros so we can keep dividing. Adding zeros after the point does not change its value.
   2.000000… ÷ 11
3. Divide step by step. At each step carry the remainder across and put a zero after it. Keep going until a remainder appears for a second time.
   11 into 2 goes 0 times, remainder 2
   11 into 20 goes 1 time, remainder 9
   11 into 90 goes 8 times, remainder 2
   The remainder 2 has appeared before, so the digits from that point repeat for ever.
4. Find the repeating block. When a remainder comes back, the same working happens again, so the same digits repeat. The repeating block is the digits that were found between the two matching remainders.
   Digits found: 0.18
   Repeating block: 18
5. Write the answer with dot notation. Put a dot over the first digit and the last digit of the repeating block (just one dot if only one digit repeats).
   211 = 0.18

Check. A different check: multiply the decimal by the denominator. Using the first few digits, 0.18181818 × 11 = 2.00000000, which is extremely close to 2. ✓

Watch out. Show the dots over the first and the last digit of the block. Writing only a few digits and “…” is not enough to show which digits repeat.

Method. A fraction whose denominator has a prime factor other than 2 or 5 gives a recurring decimal. Use long division and watch for a remainder that repeats.

1. Say what the fraction means. The fraction bar means divide. The denominator has a prime factor other than 2 or 5, so we expect the decimal to repeat.
   56 = 5 ÷ 6
2. Set up the long division 5 ÷ 6. Write the number with a decimal point and extra zeros so we can keep dividing. Adding zeros after the point does not change its value.
   5.000000… ÷ 6
3. Divide step by step. At each step carry the remainder across and put a zero after it. Keep going until a remainder appears for a second time.
   6 into 5 goes 0 times, remainder 5
   6 into 50 goes 8 times, remainder 2
   6 into 20 goes 3 times, remainder 2
   The remainder 2 has appeared before, so the digits from that point repeat for ever.
4. Find the repeating block. When a remainder comes back, the same working happens again, so the same digits repeat. The repeating block is the digits that were found between the two matching remainders.
   Digits found: 0.83
   Repeating block: 3
   Non-repeating digits: 8
5. Write the answer with dot notation. Put a dot over the first digit and the last digit of the repeating block (just one dot if only one digit repeats).
   56 = 0.83

Check. A different check: multiply the decimal by the denominator. Using the first few digits, 0.83333333 × 6 = 5.00000000, which is extremely close to 5. ✓

Watch out. Show the dots over the first and the last digit of the block. Writing only a few digits and “…” is not enough to show which digits repeat.

Method. A fraction whose denominator has a prime factor other than 2 or 5 gives a recurring decimal. Use long division and watch for a remainder that repeats.

1. Say what the fraction means. The fraction bar means divide. The denominator has a prime factor other than 2 or 5, so we expect the decimal to repeat.
   712 = 7 ÷ 12
2. Set up the long division 7 ÷ 12. Write the number with a decimal point and extra zeros so we can keep dividing. Adding zeros after the point does not change its value.
   7.000000… ÷ 12
3. Divide step by step. At each step carry the remainder across and put a zero after it. Keep going until a remainder appears for a second time.
   12 into 7 goes 0 times, remainder 7
   12 into 70 goes 5 times, remainder 10
   12 into 100 goes 8 times, remainder 4
   12 into 40 goes 3 times, remainder 4
   The remainder 4 has appeared before, so the digits from that point repeat for ever.
4. Find the repeating block. When a remainder comes back, the same working happens again, so the same digits repeat. The repeating block is the digits that were found between the two matching remainders.
   Digits found: 0.583
   Repeating block: 3
   Non-repeating digits: 58
5. Write the answer with dot notation. Put a dot over the first digit and the last digit of the repeating block (just one dot if only one digit repeats).
   712 = 0.583

Check. A different check: multiply the decimal by the denominator. Using the first few digits, 0.58333333 × 12 = 7.00000000, which is extremely close to 7. ✓

Watch out. Show the dots over the first and the last digit of the block. Writing only a few digits and “…” is not enough to show which digits repeat.

Method. A fraction whose denominator has a prime factor other than 2 or 5 gives a recurring decimal. Use long division and watch for a remainder that repeats.

1. Say what the fraction means. The fraction bar means divide. The denominator has a prime factor other than 2 or 5, so we expect the decimal to repeat.
   47 = 4 ÷ 7
2. Set up the long division 4 ÷ 7. Write the number with a decimal point and extra zeros so we can keep dividing. Adding zeros after the point does not change its value.
   4.000000… ÷ 7
3. Divide step by step. At each step carry the remainder across and put a zero after it. Keep going until a remainder appears for a second time.
   7 into 4 goes 0 times, remainder 4
   7 into 40 goes 5 times, remainder 5
   7 into 50 goes 7 times, remainder 1
   7 into 10 goes 1 time, remainder 3
   7 into 30 goes 4 times, remainder 2
   7 into 20 goes 2 times, remainder 6
   7 into 60 goes 8 times, remainder 4
   The remainder 4 has appeared before, so the digits from that point repeat for ever.
4. Find the repeating block. When a remainder comes back, the same working happens again, so the same digits repeat. The repeating block is the digits that were found between the two matching remainders.
   Digits found: 0.571428
   Repeating block: 571428
5. Write the answer with dot notation. Put a dot over the first digit and the last digit of the repeating block (just one dot if only one digit repeats).
   47 = 0.571428

Check. A different check: multiply the decimal by the denominator. Using the first few digits, 0.57142857 × 7 = 4.00000000, which is extremely close to 4. ✓

Watch out. Show the dots over the first and the last digit of the block. Writing only a few digits and “…” is not enough to show which digits repeat.

Method. A fraction whose denominator has a prime factor other than 2 or 5 gives a recurring decimal. Use long division and watch for a remainder that repeats.

1. Say what the fraction means. The fraction bar means divide. The denominator has a prime factor other than 2 or 5, so we expect the decimal to repeat.
   19 = 1 ÷ 9
2. Set up the long division 1 ÷ 9. Write the number with a decimal point and extra zeros so we can keep dividing. Adding zeros after the point does not change its value.
   1.000000… ÷ 9
3. Divide step by step. At each step carry the remainder across and put a zero after it. Keep going until a remainder appears for a second time.
   9 into 1 goes 0 times, remainder 1
   9 into 10 goes 1 time, remainder 1
   The remainder 1 has appeared before, so the digits from that point repeat for ever.
4. Find the repeating block. When a remainder comes back, the same working happens again, so the same digits repeat. The repeating block is the digits that were found between the two matching remainders.
   Digits found: 0.1
   Repeating block: 1
5. Write the answer with dot notation. Put a dot over the first digit and the last digit of the repeating block (just one dot if only one digit repeats).
   19 = 0.1

Check. A different check: multiply the decimal by the denominator. Using the first few digits, 0.11111111 × 9 = 0.99999999, which is extremely close to 1. ✓

Watch out. Show the dots over the first and the last digit of the block. Writing only a few digits and “…” is not enough to show which digits repeat.

Method. A fraction whose denominator has a prime factor other than 2 or 5 gives a recurring decimal. Use long division and watch for a remainder that repeats.

1. Say what the fraction means. The fraction bar means divide. The denominator has a prime factor other than 2 or 5, so we expect the decimal to repeat.
   322 = 3 ÷ 22
2. Set up the long division 3 ÷ 22. Write the number with a decimal point and extra zeros so we can keep dividing. Adding zeros after the point does not change its value.
   3.000000… ÷ 22
3. Divide step by step. At each step carry the remainder across and put a zero after it. Keep going until a remainder appears for a second time.
   22 into 3 goes 0 times, remainder 3
   22 into 30 goes 1 time, remainder 8
   22 into 80 goes 3 times, remainder 14
   22 into 140 goes 6 times, remainder 8
   The remainder 8 has appeared before, so the digits from that point repeat for ever.
4. Find the repeating block. When a remainder comes back, the same working happens again, so the same digits repeat. The repeating block is the digits that were found between the two matching remainders.
   Digits found: 0.136
   Repeating block: 36
   Non-repeating digits: 1
5. Write the answer with dot notation. Put a dot over the first digit and the last digit of the repeating block (just one dot if only one digit repeats).
   322 = 0.136

Check. A different check: multiply the decimal by the denominator. Using the first few digits, 0.13636364 × 22 = 3.00000000, which is extremely close to 3. ✓

Watch out. Show the dots over the first and the last digit of the block. Writing only a few digits and “…” is not enough to show which digits repeat.

Method. Write a recurring decimal as a fraction using algebra: line up the repeating parts, then subtract to cancel them.

1. Understand the aim. A recurring decimal never ends, so we cannot write it as a fraction by dividing. Instead we use algebra: we make two copies of the number whose repeating tails line up exactly, then subtract so the endless part disappears.
   0.5 = 0.555…
2. Let x equal the recurring decimal. Using a letter lets us do algebra on the number.
   x = 0.5 = 0.555…
3. Multiply by 10. 1 digit repeats, so multiplying by 10¹ = 10 moves exactly one whole block of repeating digits to the left of the point. The tail after the point is then the same as before.
   10x = 5.55…
4. Subtract the original x. Both numbers have the same infinite tail of digits after the point, so subtracting makes the tail disappear. Remember x means 1x.
   10x − x = 5.55… − 0.55…
   9x = 5
5. Solve for x. Divide both sides by the number in front of x.
   x = 59
6. Check the fraction is in its simplest form. 5 and 9 have no common factor except 1.
   x = 59

Check. A different route: divide the top by the bottom. 5 ÷ 9 gives 0.5, which matches the original. ✓

Watch out. The number you multiply by (10, 100, 1000…) depends on how many digits repeat and how many do not. Both lines must have the same repeating tail.

Method. Write a recurring decimal as a fraction using algebra: line up the repeating parts, then subtract to cancel them.

1. Understand the aim. A recurring decimal never ends, so we cannot write it as a fraction by dividing. Instead we use algebra: we make two copies of the number whose repeating tails line up exactly, then subtract so the endless part disappears.
   0.14 = 0.141414…
2. Let x equal the recurring decimal. Using a letter lets us do algebra on the number.
   x = 0.14 = 0.141414…
3. Multiply by 100. 2 digits repeat, so multiplying by 10² = 100 moves exactly one whole block of repeating digits to the left of the point. The tail after the point is then the same as before.
   100x = 14.1414…
4. Subtract the original x. Both numbers have the same infinite tail of digits after the point, so subtracting makes the tail disappear. Remember x means 1x.
   100x − x = 14.1414… − 0.1414…
   99x = 14
5. Solve for x. Divide both sides by the number in front of x.
   x = 1499
6. Check the fraction is in its simplest form. 14 and 99 have no common factor except 1.
   x = 1499

Check. A different route: divide the top by the bottom. 14 ÷ 99 gives 0.14, which matches the original. ✓

Watch out. The number you multiply by (10, 100, 1000…) depends on how many digits repeat and how many do not. Both lines must have the same repeating tail.

Method. Write a recurring decimal as a fraction using algebra: line up the repeating parts, then subtract to cancel them.

1. Understand the aim. A recurring decimal never ends, so we cannot write it as a fraction by dividing. Instead we use algebra: we make two copies of the number whose repeating tails line up exactly, then subtract so the endless part disappears.
   0.16 = 0.1666…
2. Let x equal the recurring decimal. Using a letter lets us do algebra on the number.
   x = 0.16 = 0.1666…
3. Multiply by 10. This moves the non-repeating digit (1) to the left of the point so that the repeating digits start straight after the point.
   10x = 1.66…
4. Multiply by 100. This moves one more full block of repeating digits to the left of the point, so both lines have the same tail.
   100x = 16.66…
5. Subtract. Both lines have the same repeating tail after the point, so the tail cancels.
   100x − 10x = 16 − 1
   90x = 15
6. Solve for x. Divide both sides by the number in front of x.
   x = 1590
7. Simplify the fraction. The highest common factor of 15 and 90 is 15, so divide the top and the bottom by 15.
   1590 = 16

Check. A different route: divide the top by the bottom. 1 ÷ 6 gives 0.16, which matches the original. ✓

Watch out. The number you multiply by (10, 100, 1000…) depends on how many digits repeat and how many do not. Both lines must have the same repeating tail.

Method. Write a recurring decimal as a fraction using algebra: line up the repeating parts, then subtract to cancel them.

1. Understand the aim. A recurring decimal never ends, so we cannot write it as a fraction by dividing. Instead we use algebra: we make two copies of the number whose repeating tails line up exactly, then subtract so the endless part disappears.
   0.27 = 0.2777…
2. Let x equal the recurring decimal. Using a letter lets us do algebra on the number.
   x = 0.27 = 0.2777…
3. Multiply by 10. This moves the non-repeating digit (2) to the left of the point so that the repeating digits start straight after the point.
   10x = 2.77…
4. Multiply by 100. This moves one more full block of repeating digits to the left of the point, so both lines have the same tail.
   100x = 27.77…
5. Subtract. Both lines have the same repeating tail after the point, so the tail cancels.
   100x − 10x = 27 − 2
   90x = 25
6. Solve for x. Divide both sides by the number in front of x.
   x = 2590
7. Simplify the fraction. The highest common factor of 25 and 90 is 5, so divide the top and the bottom by 5.
   2590 = 518

Check. A different route: divide the top by the bottom. 5 ÷ 18 gives 0.27, which matches the original. ✓

Watch out. The number you multiply by (10, 100, 1000…) depends on how many digits repeat and how many do not. Both lines must have the same repeating tail.

Method. Write a recurring decimal as a fraction using algebra: line up the repeating parts, then subtract to cancel them.

1. Understand the aim. A recurring decimal never ends, so we cannot write it as a fraction by dividing. Instead we use algebra: we make two copies of the number whose repeating tails line up exactly, then subtract so the endless part disappears.
   0.63 = 0.636363…
2. Let x equal the recurring decimal. Using a letter lets us do algebra on the number.
   x = 0.63 = 0.636363…
3. Multiply by 100. 2 digits repeat, so multiplying by 10² = 100 moves exactly one whole block of repeating digits to the left of the point. The tail after the point is then the same as before.
   100x = 63.6363…
4. Subtract the original x. Both numbers have the same infinite tail of digits after the point, so subtracting makes the tail disappear. Remember x means 1x.
   100x − x = 63.6363… − 0.6363…
   99x = 63
5. Solve for x. Divide both sides by the number in front of x.
   x = 6399
6. Simplify the fraction. The highest common factor of 63 and 99 is 9, so divide the top and the bottom by 9.
   6399 = 711

Check. A different route: divide the top by the bottom. 7 ÷ 11 gives 0.63, which matches the original. ✓

Watch out. The number you multiply by (10, 100, 1000…) depends on how many digits repeat and how many do not. Both lines must have the same repeating tail.

Method. Write a recurring decimal as a fraction using algebra: line up the repeating parts, then subtract to cancel them.

1. Understand the aim. A recurring decimal never ends, so we cannot write it as a fraction by dividing. Instead we use algebra: we make two copies of the number whose repeating tails line up exactly, then subtract so the endless part disappears.
   0.045 = 0.045045045…
2. Let x equal the recurring decimal. Using a letter lets us do algebra on the number.
   x = 0.045 = 0.045045045…
3. Multiply by 1000. 3 digits repeat, so multiplying by 10³ = 1000 moves exactly one whole block of repeating digits to the left of the point. The tail after the point is then the same as before.
   1000x = 45.045045…
4. Subtract the original x. Both numbers have the same infinite tail of digits after the point, so subtracting makes the tail disappear. Remember x means 1x.
   1000x − x = 45.045045… − 0.045045…
   999x = 45
5. Solve for x. Divide both sides by the number in front of x.
   x = 45999
6. Simplify the fraction. The highest common factor of 45 and 999 is 9, so divide the top and the bottom by 9.
   45999 = 5111

Check. A different route: divide the top by the bottom. 5 ÷ 111 gives 0.045, which matches the original. ✓

Watch out. The number you multiply by (10, 100, 1000…) depends on how many digits repeat and how many do not. Both lines must have the same repeating tail.