Divisibility Rules and Prime Factorization

Why can digits tell you about factors? Split a number by place value and use this idea: when one part of a sum is divisible by a positive whole number $d$, the whole sum is divisible by $d$ exactly when the other part is too. The divisible part makes equal groups of $d$ with nothing left over, so any leftover can only come from the other part.

Take $372 = 37 \times 10 + 2$. Because $10 = 2 \times 5$, the part $37 \times 10$ is divisible by $2$, by $5$ and by $10$. So you only need to check the ones digit. The ones digit $2$ is divisible by $2$, so $2$ is a factor of $372$. It is not divisible by $5$ or by $10$, so $5$ and $10$ are not factors of $372$. A whole number that is divisible by $2$ is called even, and any other whole number is called odd.

For $4$, split off the hundreds instead: $372 = 3 \times 100 + 72$. Because $100 = 4 \times 25$, the part $3 \times 100$ is divisible by $4$. So you only need to check the number made by the last two digits. Here $72 = 4 \times 18$, so $4$ is a factor of $372$.

The rules for $3$ and $9$ use every digit. Since $100 = 99 + 1$ and $10 = 9 + 1$, each hundred and each ten is a multiple of $9$ plus $1$ more:

$372 = 3 \times 100 + 7 \times 10 + 2 = (3 \times 99 + 7 \times 9) + (3 + 7 + 2)$

The first part is divisible by $9$, because $99 = 9 \times 11$, and so also by $3$, because $9 = 3 \times 3$. What is left is the digit sum, the sum of the digits: $3 + 7 + 2 = 12$. Since $12 = 3 \times 4$, the number $3$ is a factor of $372$. Since $12$ lies between $9 \times 1 = 9$ and $9 \times 2 = 18$, it is not a multiple of $9$, so $9$ is not a factor of $372$.

Longer numbers work the same way, because $1{,}000 = 999 + 1$ and $999 = 9 \times 111$, and so on for every larger place value.

You can now spot factors quickly. Factors also sort the whole numbers greater than $1$ into two kinds. A prime number has exactly two factors: $1$ and itself. A composite number has more than two factors. The number $1$ has only one factor, so it is neither prime nor composite.

For example, $7$ is prime. Dividing $7$ by $2$, $3$, $4$, $5$ or $6$ never gives a whole number (for instance, $7 \div 2 = 3.5$), so its only factors are $1$ and $7$. But $15$ is composite, because $15 = 3 \times 5$ gives it the extra factors $3$ and $5$.

The primes less than $20$ are $2$, $3$, $5$, $7$, $11$, $13$, $17$ and $19$. You will use this list to decide when a factor cannot be broken down any further.

To build a prime factorization, take out one prime factor at a time. Start with $60$. Its ones digit is $0$, so $2$ is a factor. A division gives you a multiplication fact: because $60 \div 2 = 30$, you know that $60 = 2 \times 30$. Then do the same to the quotient, and keep going:

$60 = 2 \times 30 = 2 \times 2 \times 15 = 2 \times 2 \times 3 \times 5$

Each step replaces one factor with an equal product ($30 = 2 \times 15$, then $15 = 3 \times 5$), so the value of the whole product is still $60$. You stop when every factor is prime. Here $2$, $3$ and $5$ are all on the list of primes less than $20$.

It helps to test the primes in order: $2$, then $3$, then $5$, and so on. Keep using the same prime while it still divides the quotient. The divisibility rules make the tests for $2$, $3$ and $5$ quick. For a prime with no rule in this lesson, such as $7$, divide and check whether the quotient is a whole number.

The takeaway: a factor divides a whole number with nothing left over, and divisibility rules let you find factors from the digits: the ones digit for $2$, $5$ and $10$, the last two digits for $4$, the digit sum for $3$ and $9$, and both the $2$ test and the $3$ test for $6$. To write a prime factorization, keep dividing by prime factors until every factor is prime, as in $84 = 2^2 \times 3 \times 7$, then multiply back to check. In the next unit, you will learn to find the greatest common factor and the least common multiple.