Gmat Quant Number Properties Tricks

This deck is packed with quick-reference shortcuts for handling number properties questions on the GMAT Quant section. You'll find fast divisibility rules for common divisors like 3, 4, 6, 8, 9, and 11, along with handy reminders about primes, integers, and whole numbers. There are also clever pattern-recognition tricks, such as the cyclicity of the last digit of powers of 7 and the digit-sum behavior of powers of 5, plus classic formulas for the sum of the first n integers and the sum of the first n squares.

It's a great fit if you're preparing for the GMAT and want to shave seconds off your problem-solving pace, or if you're simply brushing up on foundational number theory for any standardized math exam. Even if you already know the basics, these bite-sized cards can serve as a refresher for those moments when a divisibility rule or divisor-counting shortcut slips your mind under pressure.

To get the most out of these cards, try spacing your review sessions across several days rather than cramming everything at once, since pattern-based tricks tend to stick best with repeated, short exposures. As you study each rule, pause to think of a quick example in your head — for instance, applying the divisibility-by-11 check to a number like 1,221 — so the logic behind the shortcut becomes automatic rather than memorized.

The building blocks of number theory on the GMAT begin with classifying integers. Whole numbers include \(0, 1, 2, 3, \ldots\) while integers extend in both directions: \(\ldots, -2, -1, 0, 1, 2, \ldots\). Primes are integers greater than 1 with no divisors other than 1 and themselves; 1 is neither prime nor composite, and 2 is the smallest and only even prime. Composite numbers have a divisor other than 1 and themselves, with 4 being the smallest composite. Under 20 there are exactly eight primes: 2, 3, 5, 7, 11, 13, 17, 19. Two primes are called twin when they differ by 2, giving five pairs under 30: (3,5), (5,7), (11,13), (17,19), (29,31). The only consecutive integers that are both prime are 2 and 3, and the Fundamental Theorem of Arithmetic guarantees every integer greater than 1 a unique prime factorization up to ordering.

Many integers fall into named classifications based on the sum of their proper divisors. A number is deficient if that sum is less than itself (such as 8, whose divisors 1, 2, 4 sum to 7), abundant if the sum exceeds the number (such as 12, whose divisors sum to 16), and perfect if the sum equals the number (the classic examples are 6, 28, 496, and 8128). In fact, 28's divisors total 56, exactly twice 28. An amicable pair, such as (220, 284), is two numbers where each equals the sum of the other's proper divisors. Several named families of primes appear frequently in problems: Mersenne primes of the form \(2^p - 1\) with \(p\) itself prime (3, 7, 31, 127, 8191), Fermat primes of the form \(2^{2^n} + 1\) (3, 5, 17, 257, 65537), and Sophie Germain primes where \(p\) and \(2p+1\) are both prime (such as 2, 3, 5, 11, 23, 29).

A wider class of special integers deserves quick recognition. Harshad (or Niven) numbers are divisible by their digit sum, like 18 ÷ 9 = 2. Palindromes such as 121 and 1331 read the same forwards and backwards. Kaprekar numbers split into parts that sum back to the original when squared, so 45² = 2025 and 20 + 25 = 45. Armstrong (narcissistic) numbers equal the sum of each digit raised to the digit count, as with 153 = 1³ + 5³ + 3³. Automorphic numbers end in their own square: 5² = 25, 6² = 36, 25² = 625, 76² = 5776. Repunits, made entirely of 1's, factor in patterns such as 111111 = 3 · 7 · 11 · 13 · 37.

Parity rules are the engine behind many rapid deductions. An even number times anything is even; an odd times odd is odd; odd plus odd is even; even plus even is even; odd plus even is odd. So 17 × 23 + 9 is odd plus odd, which is even, and the expression 14n + 1 is odd for every integer n because the even term plus 1 remains odd. Consecutive integers always contain at least one even, so \(n(n+1)\) is even and \(n^2 + n\) is even. Among three consecutive integers, one is a multiple of 3 and at least one is even, so \(n(n+1)(n+2)\) is divisible by 6; \(n(n+1)(n+2)(n+3)\) is divisible by 24 = 4!. Similarly, \(n^3 - n = n(n-1)(n+1)\) is always divisible by 6. Two subtle sign traps catch many test-takers: when \(x > 1\), \(x^2 > x\), but when \(0 < x < 1\), \(x^2 < x\); and \((-3)^2 = 9\) is positive, but the expression \(-3^2\) without parentheses equals \(-9\) because the unary minus applies only to 3. A useful parity inference: if \(2x + 3y\) is even for integers \(x, y\), then \(y\) must be even, since \(3y\) has the same parity as \(y\) and the even term \(2x\) cannot flip an odd \(3y\) into an even sum.

The fastest divisibility checks on the GMAT rely on digit rules. A number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8); by 5 if its last digit is 0 or 5; by 4 if its last two digits form a multiple of 4 (such as 1532 because 32 ÷ 4 = 8); by 8 if its last three digits form a multiple of 8; by 25 if its last two digits are 00, 25, 50, or 75; and by 125 if its last three digits are one of 000, 125, 250, 375, 500, 625, 750, or 875.

Composite divisibility conditions combine the prime tests. A number is divisible by 6 if it is both even and divisible by 3; by 12 if it is divisible by both 3 and 4; and by 15 if it ends in 0 or 5 and has a digit sum divisible by 3. A common trap is assuming that divisibility by both 4 and 6 forces divisibility by 24, but in fact LCM(4, 6) = 12 is the correct threshold.

For 3, 9, 11, and 7 the digit tests are less obvious but extremely useful. A number is divisible by 3 or 9 exactly when its digit sum is divisible by 3 or 9 respectively; thus 1458 with digit sum 1+4+5+8 = 18 is divisible by 9 (and 3 and 2). For 11 the alternating sum of digits, positive at odd positions and negative at even, must be a multiple of 11, so 143 has alternating sum 1 − 4 + 3 = 0, confirming 143 = 11 × 13. The 7-test doubles the last digit and subtracts from the rest: 105 becomes 10 − (2·5) = 0, so 105 is divisible by 7, and 203 becomes 20 − 6 = 14, also divisible by 7.

Once an integer is broken into prime factors, several powerful quantities become easy to compute. To count the divisors of a number, write its prime factorization, add 1 to every exponent, and multiply the results. For 12 = 2² × 3¹ this gives (2+1)(1+1) = 6 divisors, and for 360 = 2³ × 3² × 5¹ the divisor count is (3+1)(2+1)(1+1) = 24.

The greatest common divisor of two numbers is built from the prime factors they share, each raised to the lowest exponent that appears. For 12 = 2² × 3 and 18 = 2 × 3², the shared primes 2 and 3 appear with exponents 1 and 1, giving GCD = 2 × 3 = 6. For 48 = 2⁴ × 3 and 180 = 2² × 3² × 5, the same shared primes 2 and 3 appear with minimum exponents 2 and 1, yielding GCD = 2² × 3 = 12. Two numbers are coprime when they share no prime factors, as with 18 and 35, in which case GCD = 1 and LCM = 18 · 35 = 630.

The least common multiple is built the opposite way, taking each prime to the highest exponent across all the numbers. For 12 and 18, this gives 2² × 3² = 36. For three numbers such as 12 = 2² · 3, 15 = 3 · 5, and 20 = 2² · 5, the LCM is 2² · 3 · 5 = 60. A beautiful identity is \[\mathrm{LCM}(a, b) \cdot \mathrm{GCD}(a, b) = a \cdot b,\] so LCM(a, b) = ab / GCD(a, b). To find the smallest positive integer divisible by every integer from 1 to 10, compute LCM(1, 2, …, 10), which equals 2520. Euler's totient function generalizes the coprime count: for \(n\) with factorization \(p_1^{a_1} p_2^{a_2} \cdots\), \[\varphi(n) = n \cdot \left(1 - \tfrac{1}{p_1}\right) \cdot \left(1 - \tfrac{1}{p_2}\right) \cdots.\] Applied to 36 = 2² · 3², this gives \(\varphi(36) = 36 \cdot \tfrac{1}{2} \cdot \tfrac{2}{3} = 12\).

Modular arithmetic collapses huge numbers into small, manageable remainders. The fundamental distributive property is \((a \cdot b) \bmod n = ((a \bmod n)(b \bmod n)) \bmod n\), which lets you square, multiply, and raise without expanding the original expression. If \(x\) leaves remainder 3 when divided by 7, then \(x^2\) leaves the same remainder as \(3^2 = 9\), which is 2 modulo 7. A useful fact about squares is that they are always congruent to 0 or 1 modulo 4, which is why 1234567 (whose value mod 4 is 3) cannot possibly be a perfect square.

A particularly powerful congruence is that any positive integer is congruent to the sum of its digits modulo 9. This single rule explains why the digit-sum test detects divisibility by 3 and 9. It also unlocks otherwise impossible digit-sum calculations: to find \(7^{50}\) modulo 9, note that \(7^3 = 343 \equiv 1 \pmod{9}\). Since 50 mod 3 = 2, \(7^{50} \equiv 7^2 = 49 \equiv 4 \pmod{9}\). The same logic gives 2¹⁰⁰ mod 9: since \(2^3 \equiv 8 \equiv -1\) and \(2^{99} \equiv (-1)^{33} = -1\), we get \(2^{100} \equiv -2 \equiv 7 \pmod{9}\), so the remainder is 7.

Last-digit problems are a special case of arithmetic modulo 10. Powers of 5 always end in 5 for any positive exponent, since 5¹ = 5, 5² = 25, 5³ = 125, and so on. Powers of 7 cycle through 7, 9, 3, 1 with period 4, so 7²⁵ ends in 7 because 25 mod 4 = 1. Powers of 2 cycle 2, 4, 8, 6 with period 4, so 2⁵⁰ ends in 4 since 50 mod 4 = 2. Powers of 3 (and 13, since 13 ends in 3) cycle 3, 9, 7, 1 with period 4, so 3¹⁰⁰ ends in 1 because 100 mod 4 = 0, and 13⁷⁹ ends in 7 because 79 mod 4 = 3. The pattern extends to two-digit residues for harder problems: since 37² ≡ 69 (mod 100) and 37⁴ ≡ 61 (mod 100), and 61⁵ ≡ 1 (mod 100), we get 37⁴⁰ = (37⁴)¹⁰ ≡ 1¹⁰ = 1 (mod 100), so the last two digits of 37⁴⁰ are 01. Other quick remainders: \(5^{23} \bmod 13\) equals 5 because \(5^2 \equiv -1\) and \(5^{22} \equiv 1\); \(7^{600} \bmod 5\) equals 1 because 7 ≡ 2, 2⁴ ≡ 1, and 600 mod 4 = 0; and \(3^{50} \bmod 7\) equals 2 because \(3^6 \equiv 1\) and 50 mod 6 = 2 gives \(3^2 = 9 \equiv 2\).

A factorial \(n!\) is the product \(1 \cdot 2 \cdot 3 \cdots n\), with the convention that \(0! = 1\). For reference, \(10! = 3{,}628{,}800\). Factorials connect directly to combinations through the binomial coefficient \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\); for example, \(\binom{10}{3} = \frac{10!}{3! \cdot 7!} = 120\).

To count trailing zeros in \(n!\), count factors of 5, since factors of 2 are always more abundant: \(\lfloor n/5 \rfloor + \lfloor n/25 \rfloor + \lfloor n/125 \rfloor + \cdots\). For 25! this gives \(\lfloor 25/5 \rfloor + \lfloor 25/25 \rfloor = 5 + 1 = 6\) trailing zeros. The same method finds the highest prime power dividing a factorial: the largest power of 2 in 100! is \(\lfloor 100/2 \rfloor + \lfloor 100/4 \rfloor + \lfloor 100/8 \rfloor + \lfloor 100/16 \rfloor + \lfloor 100/32 \rfloor + \lfloor 100/64 \rfloor = 50 + 25 + 12 + 6 + 3 + 1 = 97\), and the highest power of 3 in 50! is \(\lfloor 50/3 \rfloor + \lfloor 50/9 \rfloor + \lfloor 50/27 \rfloor = 16 + 5 + 1 = 22\). To estimate the number of digits in \(n!\), use \(\lfloor \log_{10}(n!) \rfloor + 1\), giving 65 digits for 50!.

Perfect squares and cubes form a familiar landscape worth memorizing: squares are \(n^2\) for \(n \geq 0\), starting 0, 1, 4, 9, 16, 25, … and cubes are \(n^3\), starting 0, 1, 8, 27, 64, 125, 216, …. Between 50 and 200 the perfect squares are 64, 81, 100, 121, 144, 169, and 196, seven in total. A square always ends in 0, 1, 4, 5, 6, or 9, so no square can end in 2, 3, 7, or 8; the last two digits of \(n^2\) also follow a fixed 22-element pattern mod 100. The factorization \(x^2 - y^2 = (x-y)(x+y)\) is the master tool for difference-of-squares problems, while \(x^2 + y^2\) does not factor over the integers (it does over the complex numbers as \((x+iy)(x-iy)\)). Pythagorean triples make \(a^2 + b^2\) a perfect square: the smallest is (3, 4, 5), and every primitive triple takes the form \(a = m^2 - n^2\), \(b = 2mn\), \(c = m^2 + n^2\) with coprime \(m > n\) of opposite parity. The irrationality of \(\sqrt{2}\) follows a classic proof by contradiction: if \(\sqrt{2} = p/q\) in lowest terms, then \(2q^2 = p^2\), forcing \(p\) even, then \(q\) even, contradicting lowest terms.

Repeating decimals convert to fractions through a single algebraic move. Let \(x\) equal the repeating block, then \(10^k x - x\) (where \(k\) is the block length) clears the repeating part. So \(0.\overline{1} = 1/9\), \(0.\overline{3} = 1/3\), \(0.1\overline{6} = 1/6\) (since \(0.1\overline{6} = 0.2 - 0.0\overline{3} = 1/5 - 1/30 = 1/6\)), \(0.\overline{142857} = 1/7\), and \(0.\overline{123} = 123/999 = 41/333\). A terminating decimal like 0.625 simply reduces: \(625/1000 = 5/8\).

Standard sum formulas cover the most common arithmetic series. The sum of the first \(n\) positive integers is \(S_n = \frac{n(n+1)}{2}\), so \(1 + 2 + \cdots + 100 = 5050\). The sum of the first \(n\) squares is \(1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}\). The sum of the first \(n\) cubes has the elegant closed form \(1^3 + 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2\), which is precisely the square of the integer sum. Finally, the sum of the first \(n\) odd numbers equals \(n^2\), so the first 50 odd numbers sum to 2500.

These formulas extend naturally to multiples and partial ranges. The sum of the first \(n\) even numbers is \(2 \cdot \frac{n(n+1)}{2} = n(n+1)\), giving 20 · 21 = 420 for the first 20 even numbers and 30 · 31 = 930 for the first 30. The sum of the first 100 multiples of 7 is \(7 \cdot \frac{100 \cdot 101}{2} = 7 \cdot 5050 = 35{,}350\). The sum of multiples of 3 from 3 to 99 has 33 terms averaging 51, giving 33 · 51 = 1683. For an arbitrary arithmetic series from \(a\) to \(b\) inclusive, the number of terms is \(b - a + 1\), the average is \((a+b)/2\), and the sum is the average times the term count. For example, \(11 + 12 + \cdots + 20\) has 10 terms averaging 15.5, summing to 155. The same averaging trick handles the median of consecutive integers: the average from \(a\) to \(b\) is always \((a+b)/2\), equal to the median.

Three classical means provide compact summaries of data. The arithmetic mean of \(v_1, \ldots, v_n\) is \(\frac{v_1 + \cdots + v_n}{n}\), so the arithmetic mean of 12, 18, 24, and 30 is \(84/4 = 21\). The geometric mean of two positive numbers \(a\) and \(b\) is \(\sqrt{ab}\); applied to 4 and 36 this gives \(\sqrt{144} = 12\). The harmonic mean is \(\frac{2ab}{a+b}\); for 2 and 8 it equals \(16/5 = 3.2\). For two equal distances traveled at speeds \(a\) and \(b\), the average round-trip speed is exactly the harmonic mean \(\frac{2ab}{a+b)}\). The median of an odd-sized ordered list is its middle value (so 3, 8, 11, 17, 25 has median 11), the mode is the most frequent value (9 in 2, 5, 5, 8, 9, 9, 9, 10), and the range is the maximum minus the minimum (15 for −3, 7, 12, 0, 5).

Percent-to-decimal conversion is a moving-the-point operation: shift the decimal two places left. Thus 12.5% becomes 0.125, and 7% of 250 equals 0.07 × 250 = 17.5. The complementary reverse calculation \(p\) is what percent of \(q\) reduces to \(p/q\) expressed as a percentage, so 15 is 20% of 75, and 25 is 12.5% of 200.

Successive percentage changes never simply add, because each change applies to the new total. A 20% increase followed by a 20% decrease takes 100 to 120 to 96, a net 4% decrease; the multiplier form is \(1.2 \times 0.8 = 0.96\). A 50% up, 50% down cycle gives \(1.5 \times 0.5 = 0.75\), a 25% decrease, and a 10% up, 10% down cycle gives \(1.1 \times 0.9 = 0.99\), a 1% decrease. A discount chain of 20% off then 10% off the sale price yields a net discount of \(1 - 0.8 \times 0.9 = 0.28\), or 28% off, and the order of operations matters when tax is added: $100 discounted 20% then taxed 8% becomes \(100 \times 0.8 \times 1.08 = 86.40\). For exponential growth, the Rule of 72 estimates doubling time as \(72/r\) years, so 6% growth doubles in roughly 12 years, and \(1.03^{10} \approx 1.3439\) reflects 3% annual growth over a decade.

Rate problems follow \(\text{speed} \times \text{time} = \text{distance}\). At 60 km/h, covering 180 km takes 3 hours. Two trains 240 km apart approaching at 40 and 80 km/h close at 120 km/h and meet in 2 hours; running in opposite directions from the same point at 50 and 70 km/h, they are \((50+70) \times 3 = 360\) km apart after 3 hours; and a faster train at 60 km/h chasing a slower train at 40 km/h from 100 km behind closes at 20 km/h, taking 5 hours to catch up. Boats on rivers combine with the current: a boat at 10 km/h in still water moves at 14 km/h downstream and 6 km/h upstream against a 4 km/h current. For a round trip of equal distances, the average speed is the harmonic form \(\frac{v^2 - u^2}{v}\), where \(v\) is the boat's still-water speed and \(u\) the current.

Work problems use the principle that combined rates add. If A finishes a job in 12 days and B in 18 days, together they complete \(1/12 + 1/18 = 5/36\) of the job per day, finishing in 36/5 = 7.2 days. If A and B together finish in 8 days and A alone takes 12, then \(1/B = 1/8 - 1/12 = 1/24\), so B alone needs 24 days. Two pipes filling a tank in 6 and 10 minutes produce \(1/6 + 1/10 = 4/15\) of the tank per minute, filling it in 15/4 = 3 minutes 45 seconds; if one pipe fills in 6 minutes and another empties in 9, the net rate \(1/6 - 1/9 = 1/18\) per minute gives a fill time of 18 minutes. Mixture problems use weighted averages: blending 6 kg of $4/kg coffee with 4 kg of $8/kg coffee gives \((6 \cdot 4 + 4 \cdot 8)/10 = 5.60\) dollars per kg. The alligation approach generalizes the idea: 10 L of 30% acid mixed with 40 L of 60% acid yields \((300 + 2400)/50 = 54\%\) acid. A classic replacement puzzle mixes 5 L of milk with 5 L of water, then removes 5 L of the resulting mixture; the milk remaining is \(5 \cdot (5/10) = 2.5\) L, since the milk content drops by the same fraction as the volume removed.