Divisibility Rules: Practice

A number is divisible by $2$ when its last digit is even ($0, 2, 4, 6, 8$). Check the final digit of each.

$64 \to \text{ends in } 4 \;(\text{even})$

So $64$ is divisible by $2$. The others end in $7$, $3$, and $1$, which are all odd.

A number is divisible by $5$ when its last digit is $0$ or $5$. Read the final digit of each.

$5{,}840 \to \text{ends in } 0$

So $5{,}840$ is divisible by $5$. The others end in $2$, $6$, and $9$.

A number is divisible by $3$ when its digit sum is a multiple of $3$. Test $27$.

$2 + 7 = 9 = 3 \times 3$

Since $9$ is a multiple of $3$, $27$ is divisible by $3$. The digit sums of the others are $7$, $10$, and $5$.

Divisible means the division leaves a remainder of $0$. Divide and check.

$20 \div 5 = 4 \text{ remainder } 0$

So $20$ is divisible by $5$, splitting into four equal groups of $5$.

An odd number has an odd last digit, so it is not divisible by $2$. Check the final digit of each.

$73 \to \text{ends in } 3 \;(\text{odd})$

So $73$ is odd. The others end in $6$, $0$, and $8$, which are all even.

Add every digit, counting the $0$ as nothing.

$1 + 2 + 0 + 7 = 10$

So the digit sum of $1{,}207$ is $10$.

A number is divisible by $3$ when its digit sum is a multiple of $3$. Test $120$.

$1 + 2 + 0 = 3$

Since $3$ is a multiple of $3$, $120$ is divisible by $3$. The digit sums of the others are $4$, $5$, and $7$.

The even digits are exactly those that make a number divisible by $2$ when they sit in the ones place.

$0, 2, 4, 6, 8$

This is the list of even digits. The other lists mix in odd digits like $1$, $3$, $5$, or $7$.

Divide $34$ by $4$ and look at the remainder.

$34 \div 4 = 8 \text{ remainder } 2$

The remainder is not $0$, so $34$ is not divisible by $4$. Being even makes it divisible by $2$, but that is not enough for $4$.

Divisible by $2$ needs an even last digit; divisible by $5$ needs a last digit of $0$ or $5$. The only digit that is both even and $0$ or $5$ is $0$.

$70 \to \text{ends in } 0$

So $70$ is divisible by both. A number divisible by both $2$ and $5$ always ends in $0$.

A number is divisible by $9$ when its digit sum is a multiple of $9$. Test $72$.

$7 + 2 = 9$

Since $9$ is a multiple of $9$, $72$ is divisible by $9$. The digit sums of the others are $7$, $11$, and $13$.

A number is divisible by $3$ when its digit sum is a multiple of $3$. Test $402$.

$4 + 0 + 2 = 6 = 3 \times 2$

Since $6$ is a multiple of $3$, $402$ is divisible by $3$. The digit sums of the others are $4$, $5$, and $8$.