Divisibility Rules - SSAT Middle Level: Quantitative

Yes, $1{,}332$ is divisible by $12$. Digit sum 9 divisible by 3, last two 32 ÷ 4 = 8, so yes for 12.

Yes, $70{,}128$ is divisible by $8$. Last three digits 128 ÷ 8 = 16, an integer, so yes.

Yes, $91{,}872$ is divisible by $3$. Digit sum 9+1+8+7+2=27, 27 ÷ 3=9, so yes.

Yes, $4{,}180$ is divisible by $11$. Alternating sum 4-1+8-0=11, divisible by 11, so yes.

Yes, $5{,}670$ is divisible by $10$. Ends with 0, meeting the rule for divisibility by 10.

Yes, $91{,}872$ is divisible by $9$. Digit sum 27, 27 ÷ 9=3, so yes.

Yes, $70{,}128$ is divisible by $4$. Last two digits 28 ÷ 4 = 7, an integer, so yes.

Yes, $2{,}736$ is divisible by $8$. Last three digits 736 ÷ 8 = 92, an integer, confirming yes.

Divisible by $4$ if the last two digits form a multiple of $4$. Since $100 \equiv 0 \pmod{4}$, divisibility by 4 depends only on the last two digits forming a multiple of 4.

Divisible by $5$ if the last digit is $0$ or $5$. This is because multiples of 5 in base 10 end in 0 or 5 to be divisible by 5.

Divisible by $6$ if it is divisible by both $2$ and $3$. As 6 = 2 × 3 and 2 and 3 are coprime, the number must satisfy both divisibility rules.

Divisible by $8$ if the last three digits form a multiple of $8$. Since $1000 \equiv 0 \pmod{8}$, divisibility by 8 relies on the last three digits being a multiple of 8.

Divisible by $9$ if the sum of its digits is divisible by $9$. This follows from $10 \equiv 1 \pmod{9}$, making the number congruent to its digit sum modulo 9.

Divisible by $10$ if the last digit is $0$. Multiples of 10 end in 0 because 10 = 2 × 5, combining rules for 2 and 5.

Divisible by $11$ if the alternating digit sum is a multiple of $11$ (including $0$). Based on $10 \equiv -1 \pmod{11}$, leading to the alternating sum being congruent to the number modulo 11.

Divisible by $12$ if it is divisible by both $3$ and $4$. Since 12 = 3 × 4 and 3 and 4 are coprime, both divisibility conditions must be met.

Yes, $742$ is divisible by $2$. The last digit of 742 is 2, which is even, confirming divisibility by 2.

No, $742$ is not divisible by $3$. Sum of digits: 7+4+2=13, not divisible by 3, so 742 is not divisible by 3.

No, $1{,}234$ is not divisible by $6$. Even but digit sum 10 not divisible by 3, so not by 6.

Divisible by $3$ if the sum of its digits is divisible by $3$. This stems from $10 \equiv 1 \pmod{3}$, so the number is congruent to the sum of its digits modulo 3.

A number is divisible by $2$ if its last digit is even ($0,2,4,6,8$). This rule works because 10 is divisible by 2, making the entire number's divisibility depend solely on the last digit's evenness.

Yes, $316$ is divisible by $4$. Last two digits 16 form 16 ÷ 4 = 4, an integer, so yes.