Divisibility Rules: Additional Practice (Set 1)

Divisibility by $8$ is the strictest test here, and any multiple of $8$ is automatically a multiple of $4$ and $2$. Test the last three digits.

$1{,}016 \to 016 = 16 = 8 \times 2$

So $1{,}016$ is divisible by $8$, hence by $4$ and $2$ as well. The others ($012, 020, 028$) are not multiples of $8$.

A number leaves the same remainder when divided by $9$ as its digit sum does. Add the digits, folding if needed.

$6 + 8 + 5 + 4 = 23, \qquad 2 + 3 = 5$

The digit sum folds down to $5$, so the remainder is $5$.

The numbers $3$ and $4$ share no common factor above $1$, so passing both tests forces divisibility by their product.

$3 \times 4 = 12$

So the number must be divisible by $12$. It need not reach $24$ or $8$; for example $12$ is divisible by $3$ and $4$ but by neither of those, and a number divisible by $3$ and $4$ need not end in $0$, so $10$ is not forced either.

Divisible by $9$ needs a digit sum that is a multiple of $9$; not divisible by $6$ means it fails the $2$ test or the $3$ test. Since a multiple of $9$ always passes the $3$ test, the only way to dodge $6$ is to be odd.

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

So $531$ is divisible by $9$ but odd, hence not by $6$. The others ($522$, $540$, $594$) are even and divisible by $9$, so they are all divisible by $6$.

Because $24 = 8 \times 3$ with $8$ and $3$ sharing no factor above $1$, test divisibility by $8$ (last three digits) and $3$ (digit sum).

$2{,}160 \to 160 = 8 \times 20 \;(\text{by } 8), \quad 2+1+6+0 = 9 \;(\text{by } 3)$

So $2{,}160$ passes both and is divisible by $24$. Of the others, $2{,}164$ and $2{,}168$ fail the $3$ test (digit sums $13$ and $17$), and $2{,}172$ passes the $3$ test but its last three digits $172$ are not a multiple of $8$.

Because $18 = 2 \times 9$ with $2$ and $9$ sharing no factor above $1$, test divisibility by $2$ (even last digit) and $9$ (digit sum a multiple of $9$).

$4{,}608 \to \text{ends in } 8 \;(\text{even}), \quad 4+6+0+8 = 18 = 9 \times 2 \;(\text{by } 9)$

So $4{,}608$ passes both and is divisible by $18$, and indeed $4{,}608 = 18 \times 256$. The others are even but fail the $9$ test (digit sums $11$, $13$, $15$).

Find the remainder when divided by $9$ using the digit sum.

$4 + 0 + 0 + 4 = 8$

The digit sum is $8$, already less than $9$, so the remainder is $8$.

Divisibility by $9$ requires the digit sum to be a multiple of $9$. Check whether $24$ qualifies.

$24 = 9 \times 2 + 6$

Since $24$ leaves remainder $6$, it is not a multiple of $9$, so the number is not divisible by $9$. (It is divisible by $3$, since $24 = 3 \times 8$.)

Because $20 = 4 \times 5$ with $4$ and $5$ sharing no factor above $1$, test divisibility by $4$ (last two digits) and $5$ (last digit $0$ or $5$).

$5{,}320 \to 20 = 4 \times 5 \;(\text{by } 4), \quad \text{ends in } 0 \;(\text{by } 5)$

So $5{,}320$ passes both and is divisible by $20$. Of the others, $5{,}310$ fails the $4$ test ($10 \div 4$ leaves remainder $2$), and $5{,}305$ and $5{,}315$ are odd.

Divisible by $4$ and by $9$, with $4$ and $9$ sharing no factor above $1$, forces divisibility by the product.

$4 \times 9 = 36$

So it is guaranteed divisible by $36$, which also covers $6$. It is not guaranteed to be divisible by $8$ (that needs a third factor of $2$) or by $24$ (which needs $8$). Among the choices, the largest guaranteed divisor is $36$.

Each candidate has a digit sum that is a multiple of $3$; the question is which one is a multiple of $3$ but not of $9$.

$4 + 5 + 1 + 2 = 12 = 3 \times 4$

The digit sum $12$ is a multiple of $3$ but not of $9$, so $4{,}512$ is divisible by $3$ and not by $9$. The others have digit sums $9$, $18$, and $18$, all multiples of $9$.

Divisibility by $8$ depends only on the last three digits, and $000$ stands for $0$, which is a multiple of $8$ ($0 = 8 \times 0$).

$\text{last three digits} = 000 = 8 \times 0$

So $1{,}000{,}000$ is divisible by $8$. Ending in a single $0$ only guarantees divisibility by $10$, not by $8$.