Divisibility Rules: Practice

Divisibility by $8$ depends only on the last three digits, since $1000 = 8 \times 125$. Test those.

$3{,}120 \to 120 = 8 \times 15$

So $3{,}120$ is divisible by $8$. The last three digits of the others are $140$, $150$, and $170$; dividing each by $8$ leaves a nonzero remainder.

A number is divisible by $3$ when its digit sum is a multiple of $3$. The known digits give $7 + 4 = 11$, so $11 + b$ must be a multiple of $3$.

$11 + 1 = 12 = 3 \times 4$

So $b = 1$ works, giving $741 = 3 \times 247$. The digits $2$, $3$, and $6$ give digit sums $13$, $14$, and $17$, none a multiple of $3$.

Divisible by $6$ means divisible by both $2$ (even last digit) and $3$ (digit sum a multiple of $3$). Test $522$.

$522 \to \text{ends in } 2 \;(\text{even}), \quad 5+2+2 = 9 = 3 \times 3$

So $522$ passes both and is divisible by $6$. The others are either odd ($527$) or fail the $3$ test ($524$ and $526$ have digit sums $11$ and $13$).

Since $6 = 2 \times 3$ and $2$ and $3$ share no factor above $1$, passing both tests forces divisibility by $6$.

$\text{divisible by } 2 \text{ and } 3 \;\Rightarrow\; \text{divisible by } 6$

It need not be divisible by $4$, $5$, or $9$. For example $6$ itself is divisible by $2$ and $3$ but by none of those.

Divisibility by $4$ depends on the last two digits, which read $\square 4$, that is the number $10b + 4$ for the missing tens digit $b$. We need $10b + 4$ to be a multiple of $4$.

$b = 2 \to 24 = 4 \times 6$

So $b = 2$ works, giving $2{,}324 = 4 \times 581$. The digits $1$, $5$, and $7$ give $14$, $54$, and $74$, none a multiple of $4$.

A number is divisible by $3$ when its digit sum is a multiple of $3$, and by $9$ when the digit sum is a multiple of $9$. We want a digit sum that is a multiple of $3$ but not $9$.

$6 + 3 + 1 + 2 = 12 = 3 \times 4$

The digit sum $12$ is a multiple of $3$ but not of $9$, so $6{,}312$ is divisible by $3$ and not by $9$. The other three have digit sums $9$, $18$, and $18$, all multiples of $9$.

A number divisible by $10$ ends in $0$, and $0$ is even and a multiple of $5$. Since $10 = 2 \times 5$, the number is a multiple of both factors.

$\text{divisible by } 10 \;\Rightarrow\; \text{divisible by } 2 \text{ and by } 5$

It is not always divisible by $4$; for example $30$ ends in $0$ but $30 \div 4$ leaves remainder $2$.

Test the last two digits of each, since $100 = 4 \times 25$.

$8{,}932 \to 32 = 4 \times 8$

So $8{,}932$ is divisible by $4$. The last two digits of the others are $14$, $26$, and $50$, none a multiple of $4$.

Divisible by $9$ means a digit sum that is a multiple of $9$; not divisible by $2$ means an odd last digit.

$531 \to 5+3+1 = 9 \;(\text{by } 9), \quad \text{ends in } 1 \;(\text{odd})$

So $531$ is divisible by $9$ but not by $2$. Of the others, $522$ and $540$ are even, and $528$ has digit sum $15$, which is not a multiple of $9$.

Divisible by $6$ means divisible by both $2$ and $3$. Test $2{,}346$.

$2{,}346 \to \text{ends in } 6 \;(\text{even}), \quad 2+3+4+6 = 15 = 3 \times 5$

So $2{,}346$ passes both tests. Of the others, $2{,}345$ and $2{,}347$ are odd, and $2{,}344$ has digit sum $13$, which fails the $3$ test.

Test the last two digits of each, since $100 = 4 \times 25$.

$7{,}028 \to 28 = 4 \times 7$

So $7{,}028$ is divisible by $4$. The last two digits of the others are $13$, $19$, and $25$, all odd and not multiples of $4$.

The combining trick works because $6$ factors into $2$ and $3$, two numbers with no common factor above $1$.

$6 = 2 \times 3, \quad 2 \text{ and } 3 \text{ share no factor above } 1$

When those factors share nothing, passing both tests forces the full product $6$ to divide the number. Not every even number is divisible by $6$; for instance $8$ is not.