One number, one prime code

The idea

Every composite number can be factorised as a product of primes, uniquely apart from the order of the factors; writing the primes in ascending order and combining equal primes as powers gives one fixed factorisation.

Three friends, one number

Asha, Ravi and Meena each break 60 into primes with a factor tree. Asha starts with 6 × 10, Ravi with 4 × 15, Meena with 2 × 30.

Will they all end up with the same primes?

• Yes, exactly the same primes
• No, each tree gives different primes
• Same primes, only in a different order

Break it your way

Pick any first split of 360. The tree finishes itself. Then sort the primes, smallest first, and see what you get.

The Fundamental Theorem of Arithmetic

Every composite number can be written as a product of primes, and this factorisation is unique, apart from the order in which the primes occur.

Put the primes in ascending order and join equal primes as powers. Then there is exactly one way to write the number:

So 2 × 3 × 5 × 7 and 3 × 5 × 7 × 2 are the same factorisation, just written in a different order.

• Number | Its prime code
• 4 | 2^2
• 253 | 11 × 23
• 360 | 2^3 × 3^2 × 5
• 32760 | 2^3 × 3^2 × 5 × 7 × 13

Notes

Every composite number is a product of primes, and this factorisation is unique apart from the order of the primes. In ascending order with powers, it is one fixed code, e.g. 32760 = 2^3 × 3^2 × 5 × 7 × 13.

Check yourself

Which is the prime factorisation of 180, written with primes in ascending order?

Ravi's factor tree of 84 ends with the primes 2, 2, 3, 7. Meera starts her tree with 7 × 12. How many 2s will her tree have?

Answer: 2

84 = 2^2 × 3 × 7 for every tree. So Meera also gets two 2s.

A number has the prime code 2^3 × 3 × 5. What is the number?

Answer: 120

8 × 3 × 5 = 120. And 120 is the only number with this prime code.

Does 6^n ever end with the digit 0, for any natural number n?

• 2 × 90. 90 is not a prime. We must keep breaking until only primes are left.
• 2 × 3 × 5 × 6. 6 is not a prime, it is 2 × 3.
• 5 × 3^2 × 2^2. These are the right primes, but the standard code lists the smallest prime first.
• 2^2 × 3^2 × 5 — correct. Yes! 4 × 9 × 5 = 180, and the primes go 2, 3, 5 in ascending order.
• Yes, for large n. A last digit 0 needs a factor 5. 6^n = 2^n × 3^n and the theorem says no other prime can sneak in.
• Never — correct. Yes! The only primes in 6^n are 2 and 3. A number ending in 0 is divisible by 5, so 5 would have to be one of its primes. By uniqueness it is not.
• Always. Check n = 1: 6 does not end in 0.