Divisibility Rules: Practice

Divisible by $2$ and $5$ together forces a last digit of $0$, so all four candidates qualify there. The deciding test is $9$: the digit sum must be a multiple of $9$.

$8 + 9 + 1 + 0 = 18 = 9 \times 2$

So $8{,}910$ passes the $9$ test (and a multiple of $9$ is automatically a multiple of $3$). The others have digit sums $19$, $20$, and $21$, none a multiple of $9$.

A number leaves the same remainder when divided by $9$ as its digit sum does, because each place value is one more than a multiple of $9$. Add the digits.

$4 + 6 + 1 + 3 = 14, \qquad 14 - 9 = 5$

The digit sum $14$ is $9 + 5$, so the remainder is $5$.

Divisible by $9$ means the digit sum is a multiple of $9$. The known digits give $3 + 7 + 2 = 12$, so $12 + b$ must be a multiple of $9$.

$12 + 6 = 18 = 9 \times 2$

So $b = 6$, giving $3{,}672 = 9 \times 408$. The next multiple of $9$ would need $12 + b = 27$, which asks for $b = 15$, not a single digit.

Because $12 = 4 \times 3$ with $4$ and $3$ sharing no factor above $1$, test divisibility by both $4$ (last two digits) and $3$ (digit sum).

$516 \to 16 = 4 \times 4 \;(\text{by } 4), \quad 5+1+6 = 12 = 3 \times 4 \;(\text{by } 3)$

So $516$ passes both and is divisible by $12$. The others fail the $4$ test, the $3$ test, or both.

Divisible by $6$ means divisible by $2$ and $3$; not divisible by $4$ means the last two digits are not a multiple of $4$.

$1{,}314 \to \text{even}, \; 1+3+1+4 = 9 \;(\text{by } 6), \quad 14 \div 4 = 3 \text{ r } 2 \;(\text{not by } 4)$

So $1{,}314$ is divisible by $6$ but not by $4$. The others ($1{,}308$, $1{,}320$, $1{,}332$) have last two digits $08$, $20$, $32$, all multiples of $4$.

Test the last three digits for $8$ and the digit sum for $9$. Both must pass.

$2{,}952 \to 952 = 8 \times 119 \;(\text{by } 8), \quad 2+9+5+2 = 18 = 9 \times 2 \;(\text{by } 9)$

So $2{,}952$ is divisible by $8$ and by $9$, hence by $72$. The others have digit sums $22$, $17$, and $25$, none a multiple of $9$.

Because $15 = 3 \times 5$ with $3$ and $5$ sharing no factor above $1$, test divisibility by both $3$ (digit sum) and $5$ (last digit $0$ or $5$).

$2{,}340 \to \text{ends in } 0 \;(\text{by } 5), \quad 2+3+4+0 = 9 \;(\text{by } 3)$

So $2{,}340$ passes both and is divisible by $15$. The others fail the $3$ test ($2{,}345$ and $2{,}350$ have digit sums $14$ and $10$) or the $5$ test ($2{,}352$ ends in $2$).

Divisible by $6$ needs an even last digit and a digit sum that is a multiple of $3$. The known digits give $5 + 2 + 8 = 15$, already a multiple of $3$, so $15 + b$ stays a multiple of $3$ only when $b$ is $0$, $3$, $6$, or $9$. Combine with "even".

$b \in \{0, 3, 6, 9\} \text{ and } b \text{ even} \;\Rightarrow\; b \in \{0, 6\}$

The smallest is $b = 0$, giving $5{,}280 = 6 \times 880$.

Divisibility by $4$ depends only on the last two digits. An even number can still fail.

$3{,}522 \to 22, \quad 22 \div 4 = 5 \text{ remainder } 2$

So $3{,}522$ is even yet not divisible by $4$. The others end in $16$, $24$, and $32$, all multiples of $4$.

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 $9$).

$4{,}500 \to \text{ends in } 0 \;(\text{even}), \quad 4+5+0+0 = 9 \;(\text{by } 9)$

So $4{,}500$ passes both and is divisible by $18$. The others fail the $9$ test (digit sums $10$, $11$, $14$) or are odd ($4{,}505$).

Find the remainder when divided by $9$ by adding the digits, folding again if needed.

$5 + 2 + 4 + 8 + 6 = 25, \qquad 2 + 5 = 7$

The digit sum folds down to $7$, which is already less than $9$, so the remainder is $7$.

A digit sum of $15$ is a multiple of $3$ but not of $9$, so the number passes the $3$ test and fails the $9$ test. Ending in $4$ means it is not divisible by $5$ or $10$.

$15 = 3 \times 5 \;\Rightarrow\; \text{digit sum divisible by } 3$

So it is definitely divisible by $3$. It fails $5$ and $10$ (last digit $4$) and fails $9$ (digit sum $15$, not a multiple of $9$).