2.1 Divisibility Rules, Prime Factorization & Factors

On the GMAT Focus Edition Quantitative Reasoning section you answer 21 Problem Solving questions in 45 minutes with no calculator. Number properties — divisibility, primes, factors, and multiples — sit in Official Guide Quantitative Review 2026–2027 Math Review 3.1, Value, Order, and Factors. Quant is arithmetic plus elementary algebra; Euclidean geometry is not tested here, and Data Sufficiency lives in Data Insights, not in this section.

The test does not reward grinding through long division. It rewards seeing the prime structure of an integer and using that structure to count factors, test divisibility, or force a number to be a square or a cube. Prime factorization is the single most useful Quant tool in this chapter: once N is written as a product of prime powers, every factor count, every later GCF/LCM, and many remainder shortcuts become mechanical.

Divisibility

An integer a is divisible by a non-zero integer b when a = b × k for some integer k. Equivalently, dividing a by b leaves remainder 0. We also say b divides a, b is a factor (divisor) of a, and a is a multiple of b.

Because there is no calculator, GMAT questions expect the digit tests below rather than repeated subtraction.

Last-digit tests (powers of 2, 5, and 10)

• 2: the last digit is even: 0, 2, 4, 6, or 8.
• 4: the last two digits form a multiple of 4. Reason: 100 is itself a multiple of 4, so only the last two digits matter. Example: 91,528 ends in 28, and 28 = 4 × 7, so 91,528 is divisible by 4.
• 8: the last three digits form a multiple of 8. Reason: 1,000 = 8 × 125. Example: 75,128 ends in 128 = 8 × 16.
• 5: the last digit is 0 or 5.
• 10: the last digit is 0.

Digit-sum tests (3 and 9)

• 3: the sum of the digits is a multiple of 3.
• 9: the sum of the digits is a multiple of 9.

These work because 10 leaves remainder 1 when divided by 3 or by 9, so a number and its digit sum share the same remainder. Example: 5,172 has digit sum 5+1+7+2 = 15, which is a multiple of 3 but not of 9, so 5,172 is divisible by 3 and not by 9.

Alternating-sum test (11)

From the right, form d0 − d1 + d2 − d3 + …. If that result is a multiple of 11 (including 0), the original number is divisible by 11. Example: 85,932 gives 2 − 3 + 9 − 5 + 8 = 11, so 85,932 is divisible by 11.

Composite divisors: use a coprime factor pair

To test divisibility by a composite M = u × v, u and v must be coprime (their GCF is 1). Then N is divisible by M if and only if N is divisible by both u and v.

• 6: divisible by 2 and 3 (GCF(2, 3) = 1).
• 12: divisible by 3 and 4 — not by 2 and 6. Trap: 18 is divisible by 2 and by 6, but 18 is not divisible by 12.
• 15: divisible by 3 and 5.
• 36: divisible by 4 and 9.

Divisibility rules at a glance

Prime numbers

A prime is an integer greater than 1 whose only positive divisors are 1 and itself. A composite has a positive divisor other than 1 and itself.

Facts the GMAT exploits

1. 1 is neither prime nor composite. It is a unit. Any claim that “the smallest prime is 1” is false.
2. 2 is the smallest prime and the only even prime. Every even integer greater than 2 is composite.
3. There are 25 primes less than 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
4. To test whether N is prime, try prime divisors p with p ≤ sqrt(N). If none divide N, N is prime.

Worked trap: Is 91 prime? sqrt(91) is less than 10, so test 2, 3, 5, and 7. Digit sum 10 is not a multiple of 3, last digit is not 0 or 5, but 91 = 7 × 13. So 91 is composite. This is a standard decoy because 91 “looks prime.”

Worked check: Is 97 prime? sqrt(97) is less than 10. 97 is not even, digit sum 16 is not a multiple of 3, last digit is not 0 or 5, and 7 × 13 = 91 while 7 × 14 = 98. No prime ≤ 9 divides 97, so 97 is prime.

Unique prime factorization

The Fundamental Theorem of Arithmetic says every integer N > 1 factors uniquely (up to order) as a product of prime powers:

N = p1^a × p2^b × p3^c × … × pk^m

All of the factor-counting formulas below are just ways of choosing those exponents.

Total number of positive divisors

A positive divisor is formed by choosing, for each prime, an exponent from 0 through its maximum. That is a+1 choices for p1, b+1 for p2, and so on.

Number of positive divisors of N = (a+1)(b+1)(c+1)…(m+1)

Worked example: 720 = 72 × 10 = (8 × 9) × (2 × 5) = 2^4 × 3^2 × 5. Number of divisors = (4+1)(2+1)(1+1) = 5 × 3 × 2 = 30.

Another example: 360 = 2^3 × 3^2 × 5 has (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24 positive divisors. Listing them is unnecessary on a timed, no-calculator item once the formula is in hand.

Distinct primes versus total prime factors

For 720 = 2^4 × 3^2 × 5:

• Distinct prime factors: 3 (namely 2, 3, and 5).
• Total prime factors counted with multiplicity: 4+2+1 = 7.

GMAT wording is precise. “How many distinct prime factors” is not the same as “how many prime factors.”

Odd divisors and even divisors

Odd divisors cannot include the prime 2. Drop 2^4 from 720 and count divisors of 3^2 × 5: (2+1)(1+1) = 6 odd divisors. Even divisors = 30 − 6 = 24.

Perfect squares and the odd-divisor-count theorem

N is a perfect square if and only if every exponent in its prime factorization is even. Squares are the only positive integers with an odd number of positive divisors, because divisors otherwise pair as (d, N/d) except when d = N/d = sqrt(N).

N is a perfect cube if and only if every exponent is a multiple of 3.

Worked example: the smallest positive integer with exactly 12 positive divisors. Factorizations of 12 as a product of integers greater than 1 give the exponent patterns p^11, p^5 q, p^3 q^2, and p q^2 r. The smallest values are 2^11 = 2,048; 2^5 × 3 = 96; 2^3 × 3^2 = 72; and 2 × 3^2 × 5 = 90. The minimum is 72.

Worked example: if 180M must be a cube and 120M must be a square, minimize positive integer M. Factor 180 = 2^2 × 3^2 × 5 and 120 = 2^3 × 3 × 5. Let M = 2^a × 3^b × 5^c. Then a+2 is a multiple of 3 and a+3 is even; b+2 is a multiple of 3 and b+1 is even; c+1 is a multiple of 3 and is even. The smallest non-negative solution is a = 1, b = 1, c = 5, so M = 2 × 3 × 5^5 = 18,750.

Keep the pairing picture in mind: if a stem tells you N has 9 positive divisors, N must be a square (of the form p^8 or p^2 q^2). Non-squares never have an odd divisor count.