Divisibility, multiples, and factors: complete GMAT guide

An integer $a$ is divisible by an integer $b$ if $a/b$ is an integer with no remainder. Equivalently, $a=bk$ for some integer $k$. For example, 84 is divisible by 7 because $84=7 \times 12$.

Divisibility is not the same as being a factor. Saying 7 divides 84 is the same as saying 7 is a factor of 84. The statements are two ways to describe the same relationship.

The rule for 11 uses alternating signs from left to right. For 2,915, compute $2-9+1-5=-11$. Since $-11$ is divisible by 11, the number is divisible by 11.

Every integer greater than 1 can be written as a product of primes in one way, apart from the order of the factors. This fact is the basis for most divisibility questions. To factor a number, divide by the smallest prime that works, then continue with the quotient.

Prime factorization makes divisibility easy. A number is divisible by another number if the first contains at least the same prime factors, with at least the same exponents. For example, 360 is divisible by 12 because $12=2^2 \times 3$, and 360 has $2^3 \times 3^2$. It is not divisible by 16 because 16 needs $2^4$ and 360 has only $2^3$.

The greatest common factor (GCF) of two integers is the largest integer that divides both. The least common multiple (LCM) is the smallest positive integer divisible by both. Use prime factorization:

• For the GCF, take the lower exponent of each shared prime.
• For the LCM, take the higher exponent of each prime that appears.

A useful relationship for two positive integers is $\text{GCF} \times \text{LCM}=a \times b$. This identity can save time when one of the values is easy to find.

Divisibility is the special case of remainder zero. For positive integers $a$ and $b$, write

where $0 \le r \lt b$. For example, $47=5 \times 9+2$, so 47 leaves remainder 2 when divided by 5.

Remainder questions often ask about an expression built from a number. If $n$ leaves remainder 3 when divided by 8, write $n=8k+3$. Then $n+5=8k+8=8(k+1)$, so $n+5$ is divisible by 8. This substitution method is faster than testing numbers.

Questions may ask whether an expression is always divisible by a certain number. Test small values to eliminate choices, but remember that a single example does not prove a general statement. To prove divisibility, factor the expression and show that the divisor is a factor.

This kind of reasoning appears in questions about consecutive integers. The product of $k$ consecutive integers is always divisible by $k!$. For example, the product of any four consecutive integers is divisible by 24.

If $n=p^a q^b r^c$, where $p,q,r$ are distinct primes, the number of positive factors of $n$ is

For $360=2^3 \times 3^2 \times 5$, the number of positive factors is $(3+1)(2+1)(1+1)=24$. This formula counts positive factors. If a question asks for factor pairs, divide the count by 2, and note that a perfect square has one unpaired factor.

Some divisibility questions use the units digit. A number is divisible by 5 if it ends in 0 or 5. It is divisible by 10 if it ends in 0. It is divisible by 2 if it ends in an even digit. These rules are quick, but they only work for divisors that depend on the last digit. For 3, 9, and 11, use the digit-sum rules instead.

A number is divisible by 4 if its last two digits form a multiple of 4. This works because 100 is divisible by 4, so the part of the number above the last two digits contributes a multiple of 100, which is already divisible by 4. The same reasoning explains why the last three digits determine divisibility by 8.

1. Check whether the question asks about a specific number, a variable expression, or a remainder.
2. Use a divisibility rule for small divisors, and prime factorization for larger or combined divisors.
3. For remainder questions, write the number as $divisor \times quotient + remainder$.
4. For "always divisible" questions, factor the expression or use consecutive integer reasoning.
5. Test small values to eliminate choices, but do not rely on one example as proof.

• Using the digit-sum rule for 9 as if it also worked for 6.
• Checking only the last digit for divisibility by 4 or 8.
• Confusing GCF and LCM.
• Forgetting that a remainder must be smaller than the divisor.
• Assuming divisibility by 4 and 6 guarantees divisibility by 24. It does not, because the two divisors share a factor of 2. The number 12 is divisible by both 4 and 6 but not by 24.
• Counting factor pairs as factors without accounting for the square root case.