Divisibility Rules: Additional Practice (Set 2)

Since $4 = 2 \times 2$ and $3$ share no factor above $1$, passing both tests forces divisibility by their product.

$4 \times 3 = 12$

So the number must be divisible by $12$. It need not be divisible by $24$, $8$, or $9$; for example $12$ itself is divisible by $4$ and $3$ but by none of those.

Because $36 = 4 \times 9$ with $4$ and $9$ sharing no factor above $1$, test divisibility by $4$ (last two digits) and $9$ (digit sum).

$1{,}908 \to 08 = 4 \times 2 \;(\text{by } 4), \quad 1+9+0+8 = 18 = 9 \times 2 \;(\text{by } 9)$

So $1{,}908$ passes both and is divisible by $36$. The others fail the $9$ test (digit sums $13$, $12$, $15$ are not multiples of $9$).

Test the last three digits for $8$ and the digit sum for $3$. We want a multiple of $8$ whose digit sum is not a multiple of $3$.

$1{,}312 \to 312 = 8 \times 39 \;(\text{by } 8), \quad 1+3+1+2 = 7 \;(\text{not by } 3)$

So $1{,}312$ is divisible by $8$ but not by $3$. The others ($1{,}320$, $1{,}344$, $1{,}368$) have digit sums $6$, $12$, and $18$, all multiples of $3$.

Because $45 = 9 \times 5$ with $9$ and $5$ sharing no factor above $1$, test divisibility by $9$ (digit sum) and $5$ (last digit $0$ or $5$).

$6{,}750 \to \text{ends in } 0 \;(\text{by } 5), \quad 6+7+5+0 = 18 = 9 \times 2 \;(\text{by } 9)$

So $6{,}750$ passes both and is divisible by $45$. The others fail the $9$ test (digit sums $23$, $17$, $20$) or the $5$ test ($6{,}752$ ends in $2$).

Divisible by $5$ forces the last digit to be $0$ or $5$. With first digit $1$ and middle digit $8$, the number is $18d$ for an end digit $d$. Divisible by $9$ needs the digit sum a multiple of $9$.

$1 + 8 + d = 9 + d, \quad d = 0 \to \text{sum } 9$

So $d = 0$ gives $180$, which ends in $0$ (passes $5$) and has digit sum $9$ (passes $9$). The choice $185$ ends in $5$ but has digit sum $14$, and $189$ does not end in $0$ or $5$.

Test the last three digits for $8$ and check $6$ (even and digit sum a multiple of $3$). We want a multiple of $8$ whose digit sum is not a multiple of $3$.

$2{,}168 \to 168 = 8 \times 21 \;(\text{by } 8), \quad 2+1+6+8 = 17 \;(\text{not by } 3)$

So $2{,}168$ is divisible by $8$ but fails the $3$ test, hence not by $6$. The others ($2{,}160$, $2{,}352$, $2{,}136$) have digit sums $9$, $12$, $12$, all multiples of $3$, and being even they are divisible by $6$.

A number leaves the same remainder when divided by $3$ as its digit sum does. Add the digits, then find the remainder.

$7 + 2 + 8 + 5 = 22, \qquad 22 = 3 \times 7 + 1$

The digit sum $22$ leaves remainder $1$ when divided by $3$, so $7{,}285$ does too.

Since $5$ and $8$ share no factor above $1$, passing both tests forces divisibility by their product.

$5 \times 8 = 40$

So the number must be divisible by $40$. It need not reach $80$; for example $40$ is divisible by $5$ and $8$ but not by $80$, and $13$ shares no factor with the requirements at all.

The numbers $2$ and $9$ share no common factor above $1$, so passing both tests forces divisibility by their product.

$2 \times 9 = 18$

So the number must be divisible by $18$. It need not reach $54$ or $72$; for example $18$ is divisible by $2$ and $9$ but by neither of those. It also need not be divisible by $15$, which shares no factor with the requirement.

Divisible by $4$ uses the last two digits $b0 = 10b$, which is a multiple of $4$ only when $b$ is even. Divisible by $9$ uses the digit sum $9 + b$, which is a multiple of $9$ only when $b = 0$ or $b = 9$.

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

So $b = 0$, giving $900 = 4 \times 225 = 9 \times 100$. The digit $9$ would satisfy the $9$ test but $990 \div 4$ leaves remainder $2$, and $3$ and $6$ fail the $9$ test.

With the same digit $b$ in the hundreds and ones places, the number is $b4b$ and its digit sum is $b + 4 + b = 2b + 4$. Divisible by $9$ needs this to be a multiple of $9$.

$2b + 4 = 18 \;\Rightarrow\; 2b = 14 \;\Rightarrow\; b = 7$

So $b = 7$ gives $747 = 9 \times 83$. The smaller choices make $2b + 4$ equal $6$, $8$, or $12$, none a multiple of $9$.

Because $72 = 8 \times 9$ with $8$ and $9$ sharing no factor above $1$, test divisibility by $8$ (last three digits) and $9$ (digit sum).

$3{,}096 \to 096 = 96 = 8 \times 12 \;(\text{by } 8), \quad 3+0+9+6 = 18 = 9 \times 2 \;(\text{by } 9)$

So $3{,}096$ passes both and is divisible by $72$. The others have digit sums $4$, $8$, $12$, none a multiple of $9$.