Home / IGCSE Mathematics 0580 / Standard form
Cambridge IGCSE Mathematics 0580 · Number

Standard form

Writing very large and very small numbers as A×10n, converting into and out of standard form, and calculating with it. Core students calculate with standard form on Paper 3 with a calculator; Extended students must also do it by hand on Paper 2.

C1.8E1.8Core and ExtendedFree lesson
What the syllabus says

C1.8 (Core)

  • 1 Use the standard form A×10n where n is a positive or negative integer and 1⩽A<10.
  • 2 Convert numbers into and out of standard form.
  • 3 Calculate with values in standard form.

Notes and examples

  • Core candidates are expected to calculate with standard form only on Paper 3.

E1.8 (Extended)

  • 1 Use the standard form A×10n where n is a positive or negative integer and 1⩽A<10.
  • 2 Convert numbers into and out of standard form.
  • 3 Calculate with values in standard form.

Quoted from the Cambridge IGCSE Mathematics 0580 syllabus for 2025 to 2027.

What standard form is

The mass of the Earth is about 5970000000000000000000000 kg. A virus may be 0.0000001 m across. Numbers like these are hard to read and easy to copy wrongly, so scientists and engineers write them in standard form:
A×10n,where 1⩽A<10 and n is an integer.

So 5970000000000000000000000=5.97×1024 and 0.0000001=1×10−7.

Both conditions matter:

  • A is at least 1 and less than 10. It has exactly one non-zero digit before the decimal point. 3.8×105 is in standard form; 38×104 and 0.38×106 are equal to it but are not.
  • n is a whole number, positive for large numbers and negative for numbers between 0 and 1.

Why it works. Each power of 10 shifts the digits one place. 103=1000, so multiplying by 103 moves every digit three places to the left. From the indices lesson, 10−2=1100, so multiplying by 10−2 moves every digit two places to the right. The power n just counts how many places the decimal point has moved.

A useful by-product: the power tells you the size at a glance. 4×109 is bigger than 9×108 however the first parts compare, because 109 is ten times 108.

Converting large numbers into standard form

  1. Place the decimal point after the first non-zero digit. This gives A.
  2. Count how many places the point has moved to get there. That number is n.
  3. A large number (10 or more) gives a positive n.

For 47000000: the first digit is 4, so A=4.7. The point has to move 7 places from the end of 47000000 to sit between the 4 and the 7. So 47000000=4.7×107.

Check by counting digits: a number with 8 digits before its decimal point is between 107 and 108, so its power is 7. The power is always one less than the number of digits in front of the point.

Converting small numbers into standard form

For a number between 0 and 1 the method is the same, but the point moves to the right, so n is negative.

For 0.00036: the first non-zero digit is 3, so A=3.6. The point moves 4 places to the right to get from 0.00036 to 3.6. So 0.00036=3.6×10−4.

A quick check: 10−4=0.0001, and 3.6×0.0001=0.00036. Another way to count: the number of zeros before the first non-zero digit, including the one before the point, is the size of the negative power. 0.00036 has four zeros in front of the 3, so the power is −4.

Converting out of standard form reverses this. For a positive power move the point right; for a negative power move it left, filling with zeros.

  • 2.05×106: move the point 6 places right to get 2050000.
  • 7.1×10−3: move the point 3 places left to get 0.0071.

Worked example 1

(a) Write 630000 in standard form. (b) Write 0.000029 in standard form. (c) Write 8.04×10−5 as an ordinary number.

  1. (a) A=6.3. The point moves 5 places, from after the last 0 to between the 6 and the 3. So 6.3×105.

  2. (b) A=2.9. The point moves 5 places to the right, so the power is −5: 2.9×10−5.

  3. (c) Move the point 5 places to the left: 8.04→0.0000804, so 0.0000804.

Answer: (a) 6.3×105 (b) 2.9×10−5 (c) 0.0000804

Check yourself

Write 0.000052 in standard form.

Answer it in the portal, where your answer is checked as you type →

Numbers that are nearly in standard form

A calculation often produces something like 34×105 or 0.6×10−3. These are correct values but not standard form, and the final answer mark needs standard form.

Adjust A by a factor of 10 and the power by 1 the opposite way, so the value does not change:

  • 34×105=3.4×10×105=3.4×106. Making A ten times smaller makes the power one bigger.
  • 0.6×10−3=6×10−1×10−3=6×10−4. Making A ten times bigger makes the power one smaller.

If you are unsure which way the power goes, write the number out in full and convert again from scratch.

Comparing and ordering. To put numbers in standard form in order, compare the powers first; only when the powers are equal do you compare the values of A. With negative powers, remember that 10−5 is smaller than 10−4, so 9×10−5<1.2×10−4.

Calculating with a calculator

On Paper 3 (Core) and Paper 4 (Extended) you calculate with standard form on a calculator. Use the ×10x key (labelled EXP or EE on some calculators) to enter the power. To enter 6.2×10−5, press 6.2, then the ×10x key, then −5.

Two things go wrong:

  • Typing ×10 and then the power key. This is fine only if you put brackets around the whole number, (6.2×10−5). The ×10x key avoids the problem, because it keeps the number together.
  • Misreading the display. A calculator may show a long answer as 7.462686567E6 or with a small raised 06. That means 7.462686567×106. Never write the answer as 7.466; that is a different number.

The question will often say "Give your answer in standard form" and "correct to 3 significant figures". Round A, and keep the power: 7.462686…×106 becomes 7.46×106.

Worked example 2 (calculator)

A grain of sand has a mass of 6.7×10−4 grams. Work out how many grains of sand there are in 5 kg of sand. Give your answer in standard form, correct to 3 significant figures.

  1. Use the same units: 5 kg is 5000 g.

  2. Number of grains =50006.7×10−4.

  3. On the calculator: 5000÷6.7×10x(−4)=7462686.57…

  4. In standard form: 7.462…×106.

  5. Correct to 3 significant figures: 7.46×106 grains.

Answer: 7.46×106

Check yourself

Use your calculator to work out (3.6×108)÷(4.8×10−3). Give your answer in standard form.

Answer it in the portal, where your answer is checked as you type →

Multiplying and dividing without a calculator Extended

Extended candidates must also calculate with standard form on Paper 2, without a calculator. The rules of indices do the work.

Multiplying. Multiply the A parts, and add the powers. Then adjust to standard form if needed.
(3×104)×(5×106)=(3×5)×104+6=15×1010=1.5×1011

Dividing. Divide the A parts, and subtract the powers.
8×1032×10−5=82×103−(−5)=4×108

This is allowed because multiplication can be done in any order: (3×104)×(5×106) is 3×5×104×106, and 104×106=1010.

Powers and roots. (2×10−3)2=22×(10−3)2=4×10−6. The power applies to both parts.

Adding and subtracting without a calculator Extended

You cannot add the A parts unless the powers are the same. 4.2×105+3.5×104 is not 7.7×109 or 7.7×105.

Two safe methods:

  1. Write both as ordinary numbers, add, and convert back: 420000+35000=455000=4.55×105.
  2. Make the powers the same: 4.2×105=42×104, so the sum is 42×104+3.5×104=45.5×104=4.55×105.

The first method is less likely to go wrong when the powers are small. The second is better when the powers are large, such as 1020, where writing the number out would take a whole line.

Worked example 3 Extended

Without using a calculator, work out (a) (6×105)×(4×10−2) and (b) 9.1×106−6×105. Give each answer in standard form.

  1. (a) Multiply the numbers: 6×4=24. Add the powers: 5+(−2)=3. So 24×103.

  2. Adjust: 24×103=2.4×104.

  3. (b) Make the powers the same: 9.1×106=91×105.

  4. 91×105−6×105=85×105.

  5. Adjust: 85×105=8.5×106.

  6. Check (b) with ordinary numbers: 9100000−600000=8500000.

Answer: (a) 2.4×104 (b) 8.5×106

Check yourself Extended

Without using a calculator, work out (2.5×106)×(6×10−2). Give your answer in standard form.

Answer it in the portal, where your answer is checked as you type →

Worked example 4 (exam style) Extended

A radio signal travels at 3×108 metres per second. A space probe is 1.8×1012 metres from the Earth. Without using a calculator, find how long the signal takes to reach the probe. Give your answer in seconds, in standard form.

  1. Time =distancespeed=1.8×10123×108.

  2. Divide the numbers: 1.8÷3=0.6. Subtract the powers: 12−8=4.

  3. 0.6×104 is not standard form. Adjust: 0.6×104=6×103.

  4. That is 6000 seconds, which is 100 minutes: a sensible size for a signal crossing the solar system.

Answer: 6×103 seconds

Key points

  • Standard form is A×10n with 1⩽A<10 and n an integer.

  • Large numbers have a positive power; numbers between 0 and 1 have a negative power.

  • The power counts how many places the decimal point moves.

  • 34×105 and 0.6×10−3 are not standard form: adjust A and change the power the opposite way.

  • On a calculator use the ×10x key and read the display as a power of 10.

  • Extended, by hand: multiply means multiply the numbers and add the powers; divide means divide the numbers and subtract the powers; add or subtract only with equal powers.

Where marks are lost

  • Leaving an answer as 15×104 or 0.75×1011 when standard form is asked for.

  • Getting the sign of the power wrong: 0.00036 is 3.6×10−4, not 3.6×104.

  • Being one out on the power: 630000 has six digits, so the power is 5.

  • Copying a calculator display such as 7.46E6 as 7.466.

  • Adding the A parts when the powers are different.

  • Adjusting the power the wrong way: making A smaller must make the power bigger.

  • Rounding the power instead of A when asked for 3 significant figures.

Related IGCSE Mathematics 0580 lessons

All IGCSE Mathematics 0580 lessons →

An original lesson written to the Cambridge IGCSE Mathematics 0580 syllabus (2025 to 2027). The worked examples and questions are our own, not Cambridge questions. Back to every IGCSE Mathematics 0580 lesson.