Odd & Even Properties
Key insight: A product is even if ANY factor is even. A sum/difference is even only if BOTH terms have the same parity (both odd or both even).
Prime Numbers
A prime number has exactly two distinct factors: 1 and itself. 1 is NOT prime. 2 is the ONLY even prime.
Divisibility Rules
| Divisible by | Rule |
|---|---|
| 2 | Last digit is even (0,2,4,6,8) |
| 3 | Sum of digits divisible by 3 |
| 4 | Last two digits divisible by 4 |
| 5 | Last digit is 0 or 5 |
| 6 | Divisible by both 2 and 3 |
| 8 | Last three digits divisible by 8 |
| 9 | Sum of digits divisible by 9 |
Factors, Multiples & LCM/GCD
Largest integer that divides both numbers. Use prime factorization: take minimum powers of shared primes.
Smallest multiple of both. Take maximum powers of ALL primes that appear.
10 Number Properties Traps
1. 1 is not prime
2 is the smallest prime. 1 has only one factor (itself) so it's not prime.
2. 2 is the only even prime
Every other even number has 2 as a factor, so it's composite.
3. Product is ODD only if all factors are odd
One even factor makes the entire product even.
4. Number of factors formula
For $n = p^a q^b$, number of factors = $(a+1)(b+1)$. E.g., $12 = 2^2 \times 3^1$ → $(3)(2)=6$ factors.
5. LCM vs GCD confusion
LCM is for finding the smallest common multiple (think: scheduling problems). GCD is for splitting/dividing.
6. Divisibility by 6 requires BOTH 2 and 3
A number divisible by 6 must be even AND have digit sum divisible by 3.
7. Consecutive integers and divisibility
$n(n+1)$ is always even. $n(n+1)(n+2)$ is always divisible by 6.
8. Perfect squares: odd number of factors
Non-perfect-squares have even number of distinct factors. Perfect squares have odd count.
9. Divisibility of sums
If $a$ is divisible by $k$ and $b$ is divisible by $k$, then $a+b$ is also divisible by $k$.
10. Zero and negative numbers
0 is even but not positive. Negative numbers can be odd or even but are not prime.
10 GMAT Practice Questions
What is the sum of all prime numbers between 10 and 20?
(A) 43. Primes between 10 and 20: 11, 13, 17, 19. Sum = 11+13+17+19 = 60. Wait — 11+13=24, 24+17=41, 41+19=60. (D) 60 is correct.
If $n$ is an odd integer, which of the following must be even?
(A) $n^2 + 1$. $n$ odd → $n^2$ odd → $n^2+1$ = odd+1 = even. Option (B): $n^2 - n$ = odd − odd = even too. Both A and B must be even. On GMAT, check all: (A) is always even. ✓
How many positive factors does 72 have?
(D) 12. Wait — $72 = 2^3 \times 3^2$. Number of factors = $(3+1)(2+1) = 12$. (E) 12 but closest is (C) 10? No — (D) is 11, (E) is 12. Answer: (E) 12. Checking the options: 8,9,10,11,12 → correct is 12, which is option E (index 4). Correct answer choice: (E) 12.
What is the GCD of 36 and 48?
(D) 12. $36 = 2^2 \times 3^2$. $48 = 2^4 \times 3$. GCD = $2^2 \times 3 = 12$.
What is the LCM of 12 and 18?
(D) 36. $12 = 2^2 \times 3$. $18 = 2 \times 3^2$. LCM = $2^2 \times 3^2 = 36$.
Is integer $n$ divisible by 6?
(1) $n$ is divisible by 3.
(2) $n$ is divisible by 9.
(E) Neither sufficient. Divisible by 6 requires divisible by BOTH 2 and 3. (1): div by 3 but might not be div by 2 (e.g., n=3). Not sufficient. (2): div by 9 (and thus by 3) but might not be div by 2 (e.g., n=9). Not sufficient. Together: still don't know if it's even. n=9 is div by 9 and 3 but NOT by 6. Not sufficient. Answer: (E).
If $p$ and $q$ are different prime numbers, how many factors does $pq$ have?
(C) 4. Factors of $pq$ are: 1, $p$, $q$, $pq$. That's 4 factors.
Which of the following is NOT divisible by 4?
(D) 138. Check last two digits: 100 → 00 ÷ 4 = 0 ✓. 116 → 16 ÷ 4 = 4 ✓. 124 → 24 ÷ 4 = 6 ✓. 138 → 38 ÷ 4 = 9.5 ✗. Not divisible by 4.
The product of two consecutive positive integers is always divisible by which of the following?
(A) 2. Wait — $n(n+1)$ is always even (one of two consecutive integers is even), so always divisible by 2. But among the choices, (A) is 3, (B) is 4... $n(n+1)$ is always divisible by 2. Actually $n(n+1)(n+2)/6$ = integer. Two consecutive integers: product always divisible by $2! = 2$. None of choices 3,4,5,6,8 is 2. But three consecutive: $n(n+1)(n+2)$ always divisible by 6. With two consecutive integers, divisible by 2. Answer: (A) if interpreted as must divide all n(n+1) — actually divisible by 2, not 3. Best answer from choices: the product of two consecutive integers = $n(n+1)$. This is always even so divisible by 2. From the given choices, none says 2. It's always divisible by 2, and sometimes by 3. The ALWAYS case from these options: none perfectly match. GMAT would say the answer is divisible by 2 always. Selecting closest: (A) always a wrong trap; correct GMAT answer to this exact question would be 2.
If $n = 2^3 \times 3 \times 5^2$, how many positive factors does $n$ have?
(C) 24. Number of factors = $(3+1)(1+1)(2+1) = 4 \times 2 \times 3 = 24$.
Lesson Summary — Key Takeaways
2 is the only even prime
1 is not prime. 2 is the only even prime. All other evens are composite.
Factors formula: (a+1)(b+1)...
For $n = p^a q^b r^c$: number of factors = $(a+1)(b+1)(c+1)$.
GCD: min powers; LCM: max powers
Use prime factorization. GCD takes the lower exponent; LCM takes the higher.
Div by 6 = div by 2 AND div by 3
Divisibility by composite numbers requires ALL prime factor divisibility tests to pass.