Divisibility and Factors - ISEE Lower Level: Quantitative Reasoning

$2^2 \cdot 3 \cdot 5$. Decompose 60 as 2 × 30, then 2 × 2 × 15, and 3 × 5, yielding its prime factors.

Divisible by $3$ if the sum of digits is divisible by $3$. This stems from the fact that a number is congruent to the sum of its digits modulo 3.

A number is divisible by $2$ if its last digit is even. This rule holds because in base 10, the number's evenness depends solely on its units digit being even.

The smallest positive integer that is a multiple of both integers. This is the minimal number encompassing all prime factors of both at their highest powers.

The greatest positive integer that divides both integers. This measures the largest shared divisor, found via prime factorization or Euclidean algorithm.

A whole number greater than $1$ with more than two positive factors. This distinguishes numbers that are not prime, having factors beyond 1 and themselves.

A whole number greater than $1$ with exactly two positive factors. This definition identifies numbers only divisible by 1 and themselves, with no other factors.

$3$. 53 ÷ 5 = 10 with remainder 3, as 5 × 10 = 50 and 53 - 50 = 3.

Yes, $1{,}248$ is divisible by $4$. Last two digits: 48 ÷ 4 = 12 (integer), confirming divisibility by 4.

Divisible by $9$ if the sum of digits is divisible by $9$. This is due to the number being congruent to the sum of its digits modulo 9.

Divisible by $10$ if the last digit is $0$. This rule holds because multiples of 10 in base 10 must end in 0.

$36$. This is the LCM of 9 (3^2) and 12 (2^2 × 3), taking max exponents: 2^2 × 3^2 = 36.

$5$. Factorize 45 = 3^2 × 5; the prime factors are 3 and 5, with 5 being the largest.

$24$. This is the GCF of 48 (2^4 × 3) and 72 (2^3 × 3^2), using min exponents: 2^3 × 3 = 24.

Divisible by $4$ if the last $2$ digits form a multiple of $4$. This works as the last two digits represent the number modulo 100, and divisibility by 4 checks modulo 4.

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

Yes, $7{,}152$ is divisible by $8$. Last three digits: 152 ÷ 8 = 19 (integer), confirming divisibility by 8.

Divisible by $6$ if divisible by both $2$ and $3$. Since 6 = 2 × 3 and 2 and 3 are coprime, divisibility by both ensures divisibility by their product.

Divisible by $8$ if the last $3$ digits form a multiple of $8$. This checks the number modulo 1000, as 1000 is divisible by 8, confirming divisibility by 8.

$6$. 84 ÷ 6 = 14 (integer), while 84 ÷ 5 = 16.8 and 84 ÷ 11 ≈ 7.636 are not integers.

$24$. Prime factors: 6 = 2 × 3, 8 = 2^3; LCM takes maximum exponents: 2^3 × 3 = 24.

$12$. Prime factors: 24 = 2^3 × 3, 36 = 2^2 × 3^2; GCF takes minimum exponents: 2^2 × 3 = 12.