Factors and multiples questions test divisibility: which numbers divide which, how many divisors a number has, and how the greatest common factor and least common multiple behave. They are a staple of GMAT Focus Quant and of Data Sufficiency, and most can be answered with almost no arithmetic.
Everything runs through prime factorization. Write 360 as 2³ × 3² × 5 and you can read off its divisors: a divisor uses each prime at most as often as 360 does, so there are (3 + 1)(2 + 1)(1 + 1) = 24 of them. The GCF takes the smaller power of each shared prime, the LCM takes the larger power of every prime, and for any two positive integers GCF × LCM equals their product.
Know the quick tests. A number is divisible by 3 or 9 when its digit sum is, by 4 when its last two digits are, and by 6 when it is divisible by both 2 and 3. For “n is divisible by 12 and 15”, n must be a multiple of their LCM, 60, not of 180.
In Data Sufficiency, the favourite trap is assuming distinct primes. Knowing that n is divisible by 4 and by 6 only tells you n is a multiple of 12, not of 24. Test the smallest case that satisfies the statement.
4 practice questions
Problem SolvingProbabilityDivisibility/Multiples/FactorsEasy
Question 1
An integer is chosen at random from the integers $1$ through $60$, inclusive. What is the probability that the integer is divisible by neither $4$ nor $6$?
A$\frac{1}{3}$
B$\frac{5}{12}$
C$\frac{7}{12}$
D$\frac{5}{8}$
E$\frac{2}{3}$
Show answer & explanation
Correct answer: E
Key Concept or Approach
Count "divisible by 4 or 6" with inclusion-exclusion (the overlap is multiples of $\text{lcm}(4,6) = 12$), then take the complement.
Step-by-Step Solution
Step 1: Multiples of 4: $60 \div 4 = 15$. Multiples of 6: $60 \div 6 = 10$. Multiples of 12: $60 \div 12 = 5$.
Step 2: Divisible by 4 or 6: $15 + 10 - 5 = 20$.
Step 3: Divisible by neither: $60 - 20 = 40$, so the probability is $\frac{40}{60} = \frac{2}{3}$.
Final Answer
Answer: E
Common trap: 7/12 - forgetting to add back the multiples of 12, so 60 - 25 = 35 and 35/60 = 7/12.
Problem SolvingDivisibility/Multiples/FactorsExponents/PowersMedium-hard
Question 2
What is the greatest integer $k$ for which $\frac{24!}{12^k}$ is an integer?
A$8$
B$10$
C$11$
D$12$
E$22$
Show answer & explanation
Correct answer: B
Key Concept or Approach
$12^k = 2^{2k} \cdot 3^k$. Count the factors of 2 and 3 in $24!$; the scarcer one (after adjusting for the exponent 2) limits $k$.
Step-by-Step Solution
Step 1: Factors of 2 in $24!$: $12 + 6 + 3 + 1 = 22$, so $2k \le 22$ and $k \le 11$.
Step 2: Factors of 3 in $24!$: $8 + 2 = 10$, so $k \le 10$.
Step 3: The greatest $k$ is $10$.
Final Answer
Answer: B
Common trap: 11 - only checking the powers of 2 because 12 contains 2 squared.
Data SufficiencyDivisibility/Multiples/FactorsMedium-hard
Question 3
If $k$ is a positive integer, is $k$ divisible by 12?
(1) $k^2$ is divisible by 12.
(2) $k^3$ is divisible by 8.
AStatement (1) ALONE is sufficient, but statement (2) ALONE is NOT sufficient.
BStatement (2) ALONE is sufficient, but statement (1) ALONE is NOT sufficient.
CBOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
DEACH statement ALONE is sufficient.
EStatements (1) and (2) TOGETHER are NOT sufficient.
Show answer & explanation
Correct answer: E
Key Concept or Approach
If a prime $p$ divides $k^n$, then $p$ divides $k$. But $p^2$ dividing $k^2$ only guarantees $p$ divides $k$, not $p^2$.
Step 2: Statement (2): $8 \mid k^3$ just means $k$ is even. $k = 2$ gives no; $k = 12$ gives yes. Not sufficient.
Step 3: Together: we still only know $6 \mid k$. $k = 6$ satisfies both ($36$ divisible by 12, $216$ divisible by 8) and is not divisible by 12; $k = 12$ is. Not sufficient.
Final Answer
Answer: E
Common trap: C: students think $12 \mid k^2$ means $12 \mid k$, or that combining with $8 \mid k^3$ supplies the missing factor of 2.
Data SufficiencyDivisibility/Multiples/FactorsNumber PropertiesVery hard
Question 4
If $n$ is a positive integer, how many positive divisors does $n$ have?
(1) $n = 2p$, where $p$ is a prime number.
(2) $n$ is divisible by 4.
AStatement (1) ALONE is sufficient, but statement (2) ALONE is NOT sufficient.
BStatement (2) ALONE is sufficient, but statement (1) ALONE is NOT sufficient.
CBOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
DEACH statement ALONE is sufficient.
EStatements (1) and (2) TOGETHER are NOT sufficient.
Show answer & explanation
Correct answer: C
Key Concept or Approach
The divisor count depends on the prime factorisation. $2p$ has 4 divisors when $p$ is an odd prime but only 3 when $p = 2$ (since $n = 4 = 2^2$).
