Factors & Multiples — Practice Set
Bank code: NUM-FAM · Section: Number Properties · 30 questions — Easy 5 · Medium 8 · Hard 12 · Extreme 5
Prime factorise first: add 1 to every exponent and multiply to count factors, and remember that factors are finite while multiples are infinite.
Every question below is live in the question bank under the ID shown — the sheet and the portal are the same questions. Attempt a level with the clock running, then check Part C.
Part A — Questions
Quantitative Comparison — the four choices are always the same, so they are not reprinted: (A) Column A is greater · (B) Column B is greater · (C) The two quantities are equal · (D) The relationship cannot be determined from the information given
Level 1 · Easy — 5 questions · ~4 min
warm-up — these must be automatic
Q1 · NUM-FAM-004 · MCQ · 45s
Which of the following is NOT a factor of 24?
(A) 1
(B) 4
(C) 6
(D) 8
(E) 9
Q2 · NUM-FAM-005 · QC · 45s
Column A: The number of positive factors of 16 Column B: The number of positive factors of 15
Q3 · NUM-FAM-007 · Numeric Entry · 50s
How many positive factors does 2^3 x 5^2 have? Enter your answer as a number.
Numeric entry — write the number.
Q4 · NUM-FAM-013 · MCQ · 50s
How many multiples of 9 are there between 1 and 100, inclusive?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 33
Q5 · NUM-FAM-015 · Select all that apply · 55s
Which of the following are factors of 48? Select all that apply.
(A) 6
(B) 7
(C) 8
(D) 10
(E) 16
Level 2 · Medium — 8 questions · ~10 min
two or three steps, one planted trap each
Q6 · NUM-FAM-017 · Numeric Entry · 90s
How many integers between 100 and 1000, inclusive, are multiples of 6 but NOT multiples of 8? Enter your answer as a number.
Numeric entry — write the number.
Q7 · NUM-FAM-020 · Select all that apply · 75s
Which of the following integers have an odd number of positive factors? Select all that apply.
(A) 16
(B) 18
(C) 25
(D) 36
(E) 50
Q8 · NUM-FAM-022 · Numeric Entry · 75s
How many integers from 1 to 200, inclusive, are multiples of both 4 and 6? Enter your answer as a number.
Numeric entry — write the number.
Q9 · NUM-FAM-023 · QC · 75s
p and q are distinct prime numbers.
Column A: The number of positive factors of pq Column B: The number of positive factors of p^2
Q10 · NUM-FAM-024 · MCQ · 80s
What is the sum of all the positive factors of 28?
(A) 15
(B) 28
(C) 55
(D) 56
(E) 84
Q11 · NUM-FAM-027 · Numeric Entry · 80s
How many integers from 1 to 500, inclusive, are multiples of 3 but NOT multiples of 9? Enter your answer as a number.
Numeric entry — write the number.
Q12 · NUM-FAM-029 · MCQ · 70s
A positive integer n is divisible by 6 and by 10. Which of the following must n be divisible by?
(A) 4
(B) 12
(C) 20
(D) 30
(E) 60
Q13 · NUM-FAM-032 · MCQ · 80s
A positive integer n has exactly 6 positive factors. If p, q and r are distinct primes, which of the following could be the prime factorisation of n?
(A) p^6
(B) p^4 x q
(C) p^2 x q^2
(D) p^2 x q
(E) p x q x r
Level 3 · Hard — 12 questions · ~20 min
where 162+ is won or lost
Q14 · NUM-FAM-039 · MCQ · 90s
How many positive integers less than or equal to 1000 have an odd number of positive factors and are divisible by 3?
(A) 10
(B) 11
(C) 16
(D) 31
(E) 333
Q15 · NUM-FAM-041 · Numeric Entry · 110s
What is the smallest positive integer that has exactly 12 positive factors? Enter your answer as a number.
Numeric entry — write the number.
Q16 · NUM-FAM-042 · MCQ · 90s
n is a positive integer and n^3 is divisible by 54. What is the smallest possible value of n?
(A) 3
(B) 6
(C) 9
(D) 18
(E) 54
Q17 · NUM-FAM-043 · QC · 100s
x is a positive integer.
Column A: The number of even positive factors of 12x Column B: The number of odd positive factors of 12x
Q18 · NUM-FAM-045 · MCQ · 100s
What is the sum of all the positive EVEN factors of 600?
(A) 49
(B) 124
(C) 1736
(D) 1859
(E) 1860
Q19 · NUM-FAM-046 · Select all that apply · 100s
Which of the following integers have more than 8 positive factors? Select all that apply.
(A) 48
(B) 64
(C) 30
(D) 72
(E) 81
Q20 · NUM-FAM-048 · Numeric Entry · 105s
N = 4 x 9 x 25. How many positive factors of N are not factors of 4, not factors of 9, and not factors of 25? Enter your answer as a number.
Numeric entry — write the number.
Q21 · NUM-FAM-049 · QC · 110s
a and b are positive integers with GCD(a, b) = 6 and LCM(a, b) = 60.
Column A: a + b Column B: 42
Q22 · NUM-FAM-050 · MCQ · 100s
How many positive integers less than 1000 are divisible by 4 but not by 6?
(A) 83
(B) 166
(C) 167
(D) 249
(E) 250
Q23 · NUM-FAM-051 · Numeric Entry · 105s
How many positive factors of 4200 are even but not divisible by 4? Enter your answer as a number.
Numeric entry — write the number.
Q24 · NUM-FAM-052 · QC · 100s
n is an integer greater than 1.
Column A: The number of positive factors of n that are less than n Column B: The number of positive factors of n that are greater than 1
Q25 · NUM-FAM-053 · Select all that apply · 100s
n is a positive integer that is a multiple of both 8 and 18. Which of the following must be true? Select all that apply.
(A) n is a multiple of 72
(B) n is a multiple of 144
(C) n is a multiple of 24
(D) n/2 is a multiple of 36
(E) n is a multiple of 48
Level 4 · Extreme — 5 questions · ~10 min
165+ — expect to need the insight, not the grind
Q26 · NUM-FAM-056 · MCQ · 120s
n = p^2 x q^2 x r, where p, q and r are distinct primes. How many pairs of positive integers (a, b) with a <= b satisfy a x b = n and GCD(a, b) = 1?
(A) 3
(B) 4
(C) 6
(D) 8
(E) 9
Q27 · NUM-FAM-057 · MCQ · 115s
What is the sum of all positive integers n <= 1000 that have exactly 5 positive factors or exactly 3 positive factors?
(A) 722
(B) 3358
(C) 4080
(D) 4081
(E) 5449
Q28 · NUM-FAM-058 · QC · 120s
p is an odd prime and k = p^2 - 1.
Column A: The number of positive factors of k Column B: 6
Q29 · NUM-FAM-059 · Select all that apply · 120s
For which of the following values of k are there exactly 15 multiples of k between 200 and 700, inclusive? Select all that apply.
(A) 31
(B) 32
(C) 33
(D) 35
(E) 36
Q30 · NUM-FAM-060 · MCQ · 125s
How many ordered pairs of positive integers (a, b) satisfy LCM(a, b) = 360?
(A) 63
(B) 75
(C) 105
(D) 210
(E) 576
Part B — Answer Key
| Q | ID | Level | Type | Answer |
|---|---|---|---|---|
| 1 | NUM-FAM-004 |
easy | MCQ | E |
| 2 | NUM-FAM-005 |
easy | QC | A |
| 3 | NUM-FAM-007 |
easy | Numeric Entry | 12 |
| 4 | NUM-FAM-013 |
easy | MCQ | C |
| 5 | NUM-FAM-015 |
easy | Select all that apply | A, C, E |
| 6 | NUM-FAM-017 |
medium | Numeric Entry | 113 |
| 7 | NUM-FAM-020 |
medium | Select all that apply | A, C, D |
| 8 | NUM-FAM-022 |
medium | Numeric Entry | 16 |
| 9 | NUM-FAM-023 |
medium | QC | A |
| 10 | NUM-FAM-024 |
medium | MCQ | D |
| 11 | NUM-FAM-027 |
medium | Numeric Entry | 111 |
| 12 | NUM-FAM-029 |
medium | MCQ | D |
| 13 | NUM-FAM-032 |
medium | MCQ | D |
| 14 | NUM-FAM-039 |
hard | MCQ | A |
| 15 | NUM-FAM-041 |
hard | Numeric Entry | 60 |
| 16 | NUM-FAM-042 |
hard | MCQ | B |
| 17 | NUM-FAM-043 |
hard | QC | A |
| 18 | NUM-FAM-045 |
hard | MCQ | C |
| 19 | NUM-FAM-046 |
hard | Select all that apply | A, D |
| 20 | NUM-FAM-048 |
hard | Numeric Entry | 20 |
| 21 | NUM-FAM-049 |
hard | QC | D |
| 22 | NUM-FAM-050 |
hard | MCQ | B |
| 23 | NUM-FAM-051 |
hard | Numeric Entry | 12 |
| 24 | NUM-FAM-052 |
hard | QC | C |
| 25 | NUM-FAM-053 |
hard | Select all that apply | A, C, D |
| 26 | NUM-FAM-056 |
extreme hard | MCQ | B |
| 27 | NUM-FAM-057 |
extreme hard | MCQ | C |
| 28 | NUM-FAM-058 |
extreme hard | QC | D |
| 29 | NUM-FAM-059 |
extreme hard | Select all that apply | B, C, D |
| 30 | NUM-FAM-060 |
extreme hard | MCQ | C |
Part C — Worked Solutions
Q1 · NUM-FAM-004 — Answer: E
Step 1 — prime factorise: 24 = 2^3 x 3. Step 2 — a factor of 24 may use at most three 2s and at most one 3, so the complete list is 1, 2, 3, 4, 6, 8, 12, 24. Step 3 — check each option against that list. A: 1 is on it. B: 4 is on it. C: 6 is on it. D: 8 is on it. E: 9 is not. Step 4 — why 9 fails: 9 = 3^2, but 24 contains only one 3, so 24/9 = 2.67 is not an integer. Check: 24/1 = 24, 24/4 = 6, 24/6 = 4, 24/8 = 3, all integers. Answer: E.
Trap. Picking A because 1 looks too small to be a factor. 1 divides every integer, so 1 is always a factor and can never answer a 'NOT a factor' question. The second slip is accepting 9 because 3 divides 24: one 3 inside 24 is not enough to supply 3^2.
Q2 · NUM-FAM-005 — Answer: A
Step 1 — Column A: 16 = 2^4, so the factor count is 4 + 1 = 5. The factors are 1, 2, 4, 8, 16. Step 2 — Column B: 15 = 3 x 5, so the factor count is (1+1)(1+1) = 4. The factors are 1, 3, 5, 15. Step 3 — compare: 5 > 4, so Column A is greater. Check: both lists were written out in full and counted. Answer: A.
Trap. Counting distinct PRIME factors instead of all factors: 16 has one prime (2) and 15 has two (3 and 5), which makes Column B look bigger and points at B. Factor count depends on the exponents through (e+1), not on how many different primes appear.
Q3 · NUM-FAM-007 — Answer: 12
Step 1 — the number is already prime factorised: 2^3 x 5^2. Step 2 — add 1 to each exponent and multiply: (3+1)(2+1) = 4 x 3 = 12. Step 3 — why it works: a factor picks its power of 2 from 2^0, 2^1, 2^2, 2^3 (4 choices) and its power of 5 from 5^0, 5^1, 5^2 (3 choices), giving 4 x 3 = 12 combinations. Check: 2^3 x 5^2 = 200, whose factors are 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 200. Twelve of them. Answer: 12.
Trap. Multiplying the exponents themselves, 3 x 2 = 6, or adding the brackets, (3+1) + (2+1) = 7. The rule is add 1 to every exponent and then MULTIPLY the results. Note also that the word 'positive' in the stem is doing real work: -1, -2, -4 and the rest divide 200 exactly too, so dropping that word would double the count to 24. A factor question that omits 'positive' has to be read twice.
Q4 · NUM-FAM-013 — Answer: C
Step 1 — smallest multiple of 9 in range: 9 x 1 = 9. Step 2 — largest multiple of 9 in range: 9 x 11 = 99, since 9 x 12 = 108 is too big. Step 3 — the multiples run from 9 x 1 to 9 x 11, so the count is 11 - 1 + 1 = 11. Faster route: floor(100/9) - floor(0/9) = 11 - 0 = 11. Check: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99. Eleven values. Answer: C.
Trap. Dropping the +1 in the counting formula: (99 - 9)/9 = 10 (option B). The same 10 comes from listing the two-digit numbers whose digits sum to 9 and missing 99, whose digits sum to 18. Dividing 100 by 9 to get 11.1 and rounding UP gives 12 (option D), but 9 x 12 = 108 is outside the range. Answering 9 (option A) reports the divisor instead of the count. Answering 33 (option E) comes from using the digit-sum-divisible-by-3 test and counting the multiples of 3: floor(100/3) = 33.
Q5 · NUM-FAM-015 — Answer: A, C, E
48 = 2^4 x 3, so every factor uses at most four 2s and at most one 3. A - 48/6 = 8, an integer; 6 = 2 x 3 fits inside 2^4 x 3. TRUE. B - 48/7 = 6.857..., not an integer; 7 is a prime that does not appear in 48. FALSE. C - 48/8 = 6, an integer; 8 = 2^3 fits inside 2^4. TRUE. D - 48/10 = 4.8, not an integer; 10 = 2 x 5 needs a 5 and 48 has none. FALSE. E - 48/16 = 3, an integer; 16 = 2^4 fits exactly. TRUE. Answer: A, C, E.
Trap. Selecting D because 48 and 10 are both even. Sharing a 2 is not enough: 10 also needs a 5, and 48 = 2^4 x 3 has no 5. Either divide, or check that the candidate's entire prime factorisation sits inside 2^4 x 3.
Q6 · NUM-FAM-017 — Answer: 113
Step 1 — count the multiples of 6 in the range: floor(1000/6) - floor(99/6) = 166 - 16 = 150. Step 2 — an integer is a multiple of both 6 and 8 exactly when it is a multiple of LCM(6, 8). Since 6 = 2 x 3 and 8 = 2^3, LCM = 2^3 x 3 = 24. Step 3 — count the multiples of 24 in the range: floor(1000/24) - floor(99/24) = 41 - 4 = 37. Step 4 — subtract the overlap from the total: 150 - 37 = 113. Check by endpoints: the multiples of 6 run 102 through 996, that is 6 x 17 through 6 x 166, and 166 - 17 + 1 = 150; the multiples of 24 run 120 through 984, that is 41 - 5 + 1 = 37. 150 - 37 = 113. Answer: 113.
Trap. Using the product 6 x 8 = 48 in place of LCM(6, 8) = 24 when counting what to exclude: that removes floor(1000/48) - floor(99/48) = 20 - 2 = 18 instead of 37, giving 150 - 18 = 132, because every multiple of 24 that is not a multiple of 48 is left in by mistake. The other slip is stopping before the exclusion and answering 150, the count of multiples of 6, or answering 37, the size of the overlap itself instead of what remains after removing it.
Q7 · NUM-FAM-020 — Answer: A, C, D
An integer has an odd number of factors exactly when it is a perfect square: factors pair as d and N/d, and only a square has a middle factor (its square root) paired with itself. A - 16 = 4^2 = 2^4, count = 4 + 1 = 5, odd. TRUE. B - 18 = 2 x 3^2, count = (1+1)(2+1) = 6, even. FALSE. C - 25 = 5^2, count = 2 + 1 = 3, odd. TRUE. D - 36 = 6^2 = 2^2 x 3^2, count = (2+1)(2+1) = 9, odd. TRUE. E - 50 = 2 x 5^2, count = (1+1)(2+1) = 6, even. FALSE. Answer: A, C, D.
Trap. Selecting B and E because each visibly contains a square (3^2 in 18, 5^2 in 50). Containing a square is not enough - EVERY exponent must be even. Both 18 and 50 carry an odd exponent on 2, so each has 6 factors, an even count.
Q8 · NUM-FAM-022 — Answer: 16
Step 1 — a number that is a multiple of both 4 and 6 is a multiple of LCM(4, 6). Since 4 = 2^2 and 6 = 2 x 3, the LCM is 2^2 x 3 = 12. Step 2 — count multiples of 12 from 1 to 200: floor(200/12) = 16, since 12 x 16 = 192 and 12 x 17 = 204 is out of range. Check: 12, 24, 36, ..., 192 is 16 values. Answer: 16.
Trap. Using the product 4 x 6 = 24 in place of the LCM, giving floor(200/24) = 8 and losing every multiple of 12 that is not a multiple of 24 (12, 36, 60, ...). Product equals LCM only when the two numbers share no factor, and 4 and 6 share a 2.
Q9 · NUM-FAM-023 — Answer: A
Step 1 — Column A: pq = p^1 x q^1 with p and q distinct, so the count is (1+1)(1+1) = 4. The factors are 1, p, q, pq. Step 2 — Column B: p^2 has count 2 + 1 = 3. The factors are 1, p, p^2. Step 3 — 4 > 3 for every choice of distinct primes, so Column A is always greater. Check with p = 2, q = 3: 6 has factors 1, 2, 3, 6 (four); 4 has 1, 2, 4 (three). With p = 5, q = 7: 35 has 1, 5, 7, 35 (four); 25 has 1, 5, 25 (three). Answer: A.
Trap. Choosing D because p and q are unknown, so pq and p^2 could be any size. Factor COUNT depends only on the exponent pattern, never on which primes are used, so the comparison is fixed at 4 versus 3. Choosing B comes from assuming a squared number automatically has more factors.
Q10 · NUM-FAM-024 — Answer: D
Step 1 — prime factorise: 28 = 2^2 x 7. Step 2 — the sum-of-factors rule uses one bracket per prime: (2^0 + 2^1 + 2^2)(7^0 + 7^1) = (1 + 2 + 4)(1 + 7). Step 3 — multiply: 7 x 8 = 56. Check by listing: 1 + 2 + 4 + 7 + 14 + 28 = 56. Correct. Answer: D.
Trap. Answering 28 (option B) by summing only the factors below 28: 1 + 2 + 4 + 7 + 14 = 28, which is what makes 28 a perfect number and makes the wrong answer look deliberate. Adding the brackets instead of multiplying gives 7 + 8 = 15 (option A), dropping the factor 1 gives 55 (option C), and counting 28 twice gives 84 (option E).
Q11 · NUM-FAM-027 — Answer: 111
Step 1 — count multiples of 3: floor(500/3) = 166, since 3 x 166 = 498. Step 2 — count multiples of 9: floor(500/9) = 55, since 9 x 55 = 495. Step 3 — every multiple of 9 is also a multiple of 3, so subtract: 166 - 55 = 111. Check on the small range 1 to 18: multiples of 3 are 3, 6, 9, 12, 15, 18 (six); multiples of 9 are 9, 18 (two); survivors are 3, 6, 12, 15 (four), and 6 - 2 = 4. The method holds. Answer: 111.
Trap. Subtracting before flooring: 500/3 - 500/9 = 166.67 - 55.56 = 111.1, which tempts 112 or a decimal answer. Floor each count first. Answering 166 means ignoring the 'but NOT multiples of 9' clause entirely.
Q12 · NUM-FAM-029 — Answer: D
Step 1 — n is a multiple of both 6 and 10, so n is a multiple of LCM(6, 10). Step 2 — 6 = 2 x 3 and 10 = 2 x 5. Take the highest power of each prime: 2^1 x 3 x 5 = 30. Step 3 — so every such n is divisible by 30. Step 4 — test the others with n = 30, which is divisible by 6 and by 10: 30/4 = 7.5, 30/12 = 2.5, 30/20 = 1.5, 30/60 = 0.5. None of A, B, C or E is forced. Check: 30/6 = 5 and 30/10 = 3, so n = 30 is a legitimate value. Answer: D.
Trap. Multiplying 6 x 10 = 60 and choosing E; that counts the shared factor 2 twice, and n = 30 satisfies the conditions without being divisible by 60. Choosing A (4) assumes n carries two 2s because 6 and 10 are both even, but the 2 in 6 and the 2 in 10 can be the same 2.
Q13 · NUM-FAM-032 — Answer: D
Apply (exponent + 1), multiplied across the primes, to each option. A - p^6 gives 6 + 1 = 7 factors. No. B - p^4 x q gives (4+1)(1+1) = 10 factors. No. C - p^2 x q^2 gives (2+1)(2+1) = 9 factors. No. D - p^2 x q gives (2+1)(1+1) = 3 x 2 = 6 factors. Yes. E - p x q x r gives (1+1)(1+1)(1+1) = 8 factors. No. Check with p = 2, q = 3: n = 2^2 x 3 = 12, whose factors are 1, 2, 3, 4, 6, 12, exactly six. Answer: D.
Trap. Choosing A because the exponent 6 matches the 6 in the stem. The exponent is not the factor count: p^6 has 7 factors. Exactly 6 factors needs the (e+1) values to multiply to 6, i.e. 6 = 6 x 1 (p^5) or 6 = 3 x 2 (p^2 x q). The p^5 form is valid too but is not offered here.
Q14 · NUM-FAM-039 — Answer: A
Step 1 — an integer has an odd number of factors exactly when it is a perfect square, because factors pair as d and N/d and only a square leaves its square root unpaired. Step 2 — so the question asks for the perfect squares up to 1000 that are also divisible by 3. If n^2 is divisible by 3 then 3 divides n, since 3 is prime. So n = 3k and n^2 = 9k^2. Step 3 — require 9k^2 <= 1000, that is k^2 <= 111.1, so k <= 10 because 10^2 = 100 and 11^2 = 121. Step 4 — k runs 1 to 10, giving 10 values. Check by listing: 9, 36, 81, 144, 225, 324, 441, 576, 729, 900. Ten of them, and the next one, 33^2 = 1089, exceeds 1000. Answer: A.
Trap. Ignoring the divisibility condition and counting all 31 perfect squares up to 1000 (option D). Rounding sqrt(1000/9) = 10.54 UP to 11 (option B) admits 33^2 = 1089, which is over 1000. Reading 'odd number of factors' as 'the number is odd' and counting the odd perfect squares up to 1000 gives 16 (option C). Dropping the factor-count condition and counting the multiples of 3 up to 1000 gives floor(1000/3) = 333 (option E).
Q15 · NUM-FAM-041 — Answer: 60
Step 1 — exactly 12 factors means the (exponent + 1) values multiply to 12. Every way of writing 12 gives an exponent pattern: 12 = 12 gives p^11, smallest 2^11 = 2048 12 = 6 x 2 gives p^5 x q, smallest 2^5 x 3 = 96 12 = 4 x 3 gives p^3 x q^2, smallest 2^3 x 3^2 = 72 12 = 3 x 2 x 2 gives p^2 x q x r, smallest 2^2 x 3 x 5 = 60 Step 2 — in each pattern give the largest exponent to the smallest prime, as done above. Step 3 — compare the four candidates: 2048, 96, 72, 60. The smallest is 60. Check: 60 = 2^2 x 3 x 5, so the count is (2+1)(1+1)(1+1) = 12, and the factors are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. Twelve of them. Answer: 60.
Trap. Stopping at the first pattern that works, usually 2^3 x 3^2 = 72, without testing the three-prime pattern 2^2 x 3 x 5 = 60. Spreading exponents over more small primes generally gives a smaller number. Answering 96 comes from 2^5 x 3, and 2048 from insisting on a single prime.
Q16 · NUM-FAM-042 — Answer: B
Step 1 — prime factorise the divisor: 54 = 2 x 3^3. Step 2 — write n = 2^a x 3^b x (other primes). Then n^3 = 2^(3a) x 3^(3b) x ... Step 3 — for 2 x 3^3 to divide n^3 we need 3a >= 1 and 3b >= 3. Since a is a non-negative integer, 3a >= 1 forces a >= 1; and 3b >= 3 forces b >= 1. Step 4 — the smallest such n is 2^1 x 3^1 = 6. Check: 6^3 = 216 and 216/54 = 4, an integer. And 3^3 = 27, with 27/54 not an integer, so 3 is too small. Answer: B.
Trap. Answering 3 (option A) by matching only the 3^3 inside 54 and forgetting its single 2: 27/54 is not an integer. Answering 9 (option C) keeps 3^2 - taking the 3^2 visible in 54 = 2 x 27 as the requirement on n - and again drops the 2. Answering 18 (option D) demands that same 3^2 and remembers the 2, but 3^1 is already enough because cubing triples the exponent. Answering 54 (option E) assumes n itself must be divisible by 54, while the condition is on n^3.
Q17 · NUM-FAM-043 — Answer: A
Use the capsule's split: odd factors come from dropping all the 2s, and every factor is an odd factor times a power of 2. Step 1 — write 12x = 2^a x m with m odd. Since 12 = 2^2 x 3 already contributes two 2s, a >= 2 for every x. Step 2 — let t be the number of factors of m, that is the number of ODD factors of 12x. Column B = t. Step 3 — an even factor is 2^i x d with 1 <= i <= a and d a factor of m, so Column A = a x t. Step 4 — a >= 2 gives Column A = a x t >= 2t > t = Column B for every positive integer x. Check x = 1: 12 = 2^2 x 3 has factors 1, 2, 3, 4, 6, 12; even ones are 2, 4, 6, 12 (four), odd ones are 1, 3 (two). Check x = 25: 300 = 2^2 x 3 x 5^2 has 18 factors, 6 odd and 12 even. Column A wins both times. Answer: A.
Trap. Choosing D because x is unknown and 12x could be anything. The ratio is locked: every odd factor is matched by a even ones, and 12 guarantees a >= 2, so Column A is at least twice Column B. Choosing C comes from imagining the factors split evenly into odd and even, which happens only when a = 1 - impossible here.
Q18 · NUM-FAM-045 — Answer: C
Step 1 — prime factorise: 600 = 8 x 75 = 2^3 x 3 x 5^2. Step 2 — a factor of 600 is even exactly when its power of 2 is at least 1, so the 2-bracket loses its 1 and runs 2 + 4 + 8 while the odd brackets are unchanged: (2 + 4 + 8) x (1 + 3) x (1 + 5 + 25). Step 3 — evaluate the brackets: 14 x 4 x 31. Step 4 — multiply: 14 x 4 = 56, and 56 x 31 = 1736. Check the other way: the full sum-of-factors product is (1 + 2 + 4 + 8) x 4 x 31 = 15 x 4 x 31 = 1860, and the ODD factors sum to (1) x (1 + 3) x (1 + 5 + 25) = 4 x 31 = 124. Then 1860 - 124 = 1736. Correct. Answer: C. 1736.
Trap. Answering 1860 (option E) gives the sum of ALL the factors, forgetting to strip the odd ones by dropping the 1 from the 2-bracket. Answering 124 (option B) sums the ODD factors instead, the exact complement. Answering 1859 (option D) removes only the factor 1 from the full sum, treating it as the single odd factor to discard and missing 3, 5, 15, 25 and 75. Answering 49 (option A) adds the brackets instead of multiplying them: 14 + 4 + 31 = 49.
Q19 · NUM-FAM-046 — Answer: A, D
Count each with the (exponent + 1) rule. A - 48 = 2^4 x 3, count = (4+1)(1+1) = 10. 10 > 8. TRUE. B - 64 = 2^6, count = 6 + 1 = 7. Not more than 8. FALSE. C - 30 = 2 x 3 x 5, count = (1+1)(1+1)(1+1) = 8. Exactly 8 is not MORE than 8. FALSE. D - 72 = 2^3 x 3^2, count = (3+1)(2+1) = 12. 12 > 8. TRUE. E - 81 = 3^4, count = 4 + 1 = 5. FALSE. Answer: A, D.
Trap. Selecting C: 30 has exactly 8 factors, and the stem says MORE than 8, so it fails on the boundary. Selecting B because 64 is the largest option: size is irrelevant, and 64 = 2^6 has only 7 factors, fewer than the much smaller 30. One prime raised high yields surprisingly few factors.
Q20 · NUM-FAM-048 — Answer: 20
Step 1 — N = 4 x 9 x 25 = 900 = 2^2 x 3^2 x 5^2, so N has (2+1)(2+1)(2+1) = 27 positive factors. Step 2 — list the factors of the three parts: 4 gives 1, 2, 4; 9 gives 1, 3, 9; 25 gives 1, 5, 25. Step 3 — take their union, counting the shared 1 only once: 1, 2, 4, 3, 9, 5, 25. That is 7 distinct integers, and every one of them divides 900. Step 4 — subtract: 27 - 7 = 20. Check a sample of the 20 survivors: 6 divides 900 but divides none of 4, 9, 25; so do 10, 15, 12, 45, 100, 900. Answer: 20.
Trap. Assuming a factor of a product must be a factor of one of the parts, which would make the answer 0: 6 divides 4 x 9 = 36 and 15 divides 9 x 25 = 225, yet 6 divides neither 4 nor 9. The second slip is adding the three factor lists as 3 + 3 + 3 = 9 and answering 27 - 9 = 18, double-counting the 1 that all three share.
Q21 · NUM-FAM-049 — Answer: D
Step 1 — use GCD x LCM = a x b: 6 x 60 = 360, so ab = 360. Step 2 — each of a and b is a multiple of 6 and a factor of 60, so each is one of 6, 12, 30, 60. Step 3 — find the pairs whose product is 360 with GCD exactly 6: 6 and 60: product 360, GCD 6, LCM 60. Valid, a + b = 66. 12 and 30: product 360, GCD 6, LCM 60. Valid, a + b = 42. (Swapping a and b changes nothing.) Step 4 — a + b can be 66, which is greater than 42, or 42, which equals 42. Two different comparisons, so nothing is settled. Answer: D.
Trap. Finding 12 and 30 first, computing a + b = 42, and choosing C. Fixing the GCD and LCM does not fix the pair: 6 and 60 also meet both conditions and give 66. On a QC that supplies GCD and LCM, list every valid pair before comparing.
Q22 · NUM-FAM-050 — Answer: B
Step 1 — count multiples of 4 below 1000: floor(999/4) = 249, the largest being 996 = 4 x 249. Step 2 — remove those also divisible by 6. A number divisible by both 4 and 6 is divisible by LCM(4, 6) = 12. Step 3 — count multiples of 12 below 1000: floor(999/12) = 83, the largest being 996 = 12 x 83. Step 4 — subtract: 249 - 83 = 166. Check on 1 to 24: multiples of 4 are 4, 8, 12, 16, 20, 24 (six); multiples of 12 are 12, 24 (two); survivors are 4, 8, 16, 20 (four), and 6 - 2 = 4. The method holds. Answer: B.
Trap. Subtracting the 166 multiples of 6 from the 249 multiples of 4 to get 83 (option A). Only numbers divisible by BOTH should be removed, and those are the multiples of 12, not of 6. Reading 'less than 1000' as 'at most 1000' admits 1000 and gives 250 - 83 = 167 (option C); ignoring the exclusion entirely gives 249 (option D), and making both slips at once - counting up to and including 1000 and never excluding anything - gives floor(1000/4) = 250 (option E).
Q23 · NUM-FAM-051 — Answer: 12
Step 1 — prime factorise: 4200 = 42 x 100 = (2 x 3 x 7)(2^2 x 5^2) = 2^3 x 3 x 5^2 x 7. Step 2 — total factors: (3+1)(1+1)(2+1)(1+1) = 4 x 2 x 3 x 2 = 48. Step 3 — odd factors = drop all the 2s and count the factors of 3 x 5^2 x 7: (1+1)(2+1)(1+1) = 12. Even factors = total minus odd = 48 - 12 = 36. Step 4 — 'even but not divisible by 4' means the factor carries exactly one 2, that is 2^1 times an odd factor. Step 5 — there are 12 odd factors, so there are 12 such numbers. Check the same way from the other side: factors with 2^2 or 2^3 number 2 x 12 = 24, and 36 - 24 = 12. Sample survivors: 2, 6, 10, 14, 30, 50, 70, 150, 350, 1050. Each is even and each leaves remainder 2 on division by 4. Answer: 12.
Trap. Answering 36 by stopping at 'even factors = total minus odd' and never applying the 'not divisible by 4' filter. Answering 24 counts the opposite slice, the even factors that ARE divisible by 4 (those with 2^2 or 2^3). Answering 48 reports the total factor count and ignores both conditions.
Q24 · NUM-FAM-052 — Answer: C
Step 1 — let n have d positive factors in total. Step 2 — Column A takes the full factor list and removes exactly one member, n itself, so Column A = d - 1. Step 3 — Column B takes the same list and removes exactly one member, 1, so Column B = d - 1. Step 4 — since n > 1, the numbers 1 and n are distinct, so each column drops exactly one element. They are always equal. Check with n = 12 (factors 1, 2, 3, 4, 6, 12): below 12 gives 1, 2, 3, 4, 6, five; above 1 gives 2, 3, 4, 6, 12, five. With the prime n = 7 (factors 1, 7): one each. With the square n = 9 (factors 1, 3, 9): two each. Answer: C.
Trap. Choosing D because n is unknown and the two columns describe different SETS of numbers. They are different sets but always the same SIZE: each is the full factor list minus one element. Trying a perfect square such as 9, in the hope the repeated square root breaks the symmetry, also fails - 3 is counted once on each side.
Q25 · NUM-FAM-053 — Answer: A, C, D
Step 1 — a common multiple of 8 = 2^3 and 18 = 2 x 3^2 must carry 2^3 and 3^2, so n is a multiple of LCM(8, 18) = 2^3 x 3^2 = 72. Every such n is 72m, and 'must be true' means true for every m; n = 72 is the test case that kills the false options. A - n = 72m by Step 1. TRUE. B - 144 = 2^4 x 3^2 needs a fourth 2, which n does not guarantee: 72 is not a multiple of 144. FALSE. C - 24 = 2^3 x 3 divides 72, so it divides every multiple of 72. TRUE. D - n/2 = 36m, a multiple of 36 for every m. TRUE. E - 48 = 2^4 x 3 again needs a fourth 2: 72/48 = 1.5. FALSE. Answer: A, C, D.
Trap. Multiplying 8 x 18 = 144 instead of taking the LCM and selecting B; the shared factor of 2 is counted twice that way. Selecting E for the same reason, since 48 also demands 2^4 while n guarantees only 2^3. In both cases n = 72 is the single counterexample that settles it.
Q26 · NUM-FAM-056 — Answer: B
Step 1 — GCD(a, b) = 1 means a and b share no prime. Since a x b = n, each prime power of n must go ENTIRELY to a or entirely to b: p^2 cannot be split, and neither can q^2 or r. Step 2 — building a pair is therefore three independent two-way choices, one per distinct prime: 2 x 2 x 2 = 8 ORDERED pairs (a, b). Step 3 — the condition a <= b keeps one of each mirror pair. No pair has a = b, since n is not a perfect square (r has exponent 1), so the 8 ordered pairs form 4 mirror pairs. Step 4 — 8/2 = 4. Check with p = 2, q = 3, r = 7: n = 4 x 9 x 7 = 252. The coprime pairs with a <= b are (1, 252), (4, 63), (7, 36) and (9, 28). Four of them. Answer: B.
Trap. Answering 8 (option D) by counting ordered pairs and ignoring that a <= b identifies (a, b) with (b, a). Answering 3 (option A) comes from discarding (1, n) as trivial, but 1 is coprime to everything and that pair is legitimate. Answering 6 (option C) comes from counting the 8 ordered pairs and then striking out both trivial ones, (1, n) and (n, 1), to get 2^3 - 2 = 6; that both forgets to halve for a <= b and wrongly rejects (1, n), which is a genuine coprime pair. Answering 9 (option E) drops the GCD condition altogether: n has (2+1)(2+1)(1+1) = 18 factors, which form 9 unordered pairs with product n, only 4 of which are coprime.
Q27 · NUM-FAM-057 — Answer: C
Step 1 — the factor count is a product of (exponent + 1) terms. 5 is prime, so it can only be a single bracket of 5: n = p^4. Likewise 3 is prime, so n = p^2. Step 2 — fourth powers of primes up to 1000: 2^4 = 16, 3^4 = 81, 5^4 = 625. Next is 7^4 = 2401, too big. Sum of this family: 16 + 81 + 625 = 722. Step 3 — squares of primes up to 1000 need p <= 31, since 31^2 = 961 and 37^2 = 1369 > 1000. The primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, giving 4 + 9 + 25 + 49 + 121 + 169 + 289 + 361 + 529 + 841 + 961. Running total: 4 + 9 = 13, +25 = 38, +49 = 87, +121 = 208, +169 = 377, +289 = 666, +361 = 1027, +529 = 1556, +841 = 2397, +961 = 3358. Step 4 — the two families cannot overlap, since p^4 has 5 factors and p^2 has 3. Total = 722 + 3358 = 4080. Check: 16 has factors 1, 2, 4, 8, 16 (five) and 961 has 1, 31, 961 (three). Both belong. Answer: C.
Trap. Answering 722 (option A) or 3358 (option B) by solving only one of the two families and stopping. Adding 1 to the total on the view that 1 qualifies gives 4081 (option D); 1 has exactly one factor, not three. Taking 'primes up to sqrt(1000) = 31.6' as 'primes up to 37' and including 37^2 = 1369 gives 5449 (option E).
Q28 · NUM-FAM-058 — Answer: D
Step 1 — factor the expression: k = p^2 - 1 = (p - 1)(p + 1), the two even numbers on either side of p. Step 2 — test p = 3: k = 8 = 2^3, with 3 + 1 = 4 factors. 4 < 6, so Column B is greater here. Step 3 — test p = 5: k = 24 = 2^3 x 3, with (3+1)(1+1) = 8 factors. 8 > 6, so Column A is greater here. Step 4 — two opposite outcomes, so the relationship is not determined. Check a third value, p = 7: k = 48 = 2^4 x 3 with (4+1)(1+1) = 10 factors, again above 6; but p = 3 has already broken the pattern. Answer: D.
Trap. Testing p = 5, 7 and 11 (factor counts 8, 10 and 16, all above 6) and settling on A. The smallest odd prime is the exception: p = 3 gives p^2 - 1 = 8, a pure power of 2 with only 4 factors. On QC, always test the smallest legal value before committing.
Q29 · NUM-FAM-059 — Answer: B, C, D
Use count = floor(700/k) - floor(199/k), which counts the multiples of k up to 700 and removes those below 200. A - k = 31: floor(700/31) = 22 (31 x 22 = 682) and floor(199/31) = 6 (31 x 6 = 186). 22 - 6 = 16. FALSE. B - k = 32: floor(700/32) = 21 (32 x 21 = 672) and floor(199/32) = 6 (32 x 6 = 192). 21 - 6 = 15. TRUE. C - k = 33: floor(700/33) = 21 (33 x 21 = 693) and floor(199/33) = 6 (33 x 6 = 198). 21 - 6 = 15. TRUE. D - k = 35: floor(700/35) = 20 (35 x 20 = 700, included) and floor(199/35) = 5 (35 x 5 = 175). 20 - 5 = 15. TRUE. E - k = 36: floor(700/36) = 19 (36 x 19 = 684) and floor(199/36) = 5 (36 x 5 = 180). 19 - 5 = 14. FALSE. Check D by endpoints: the multiples run 210, 245, ..., 700, that is 35 x 6 through 35 x 20, and 20 - 6 + 1 = 15. Answer: B, C, D.
Trap. Estimating the count as (700 - 200)/k, the length of the range over k: that gives 16.1 at k = 31 and 13.9 at k = 36, so the estimate wrongly admits A, whose true count is 16, and wrongly rejects nothing, since its low reading at k = 36 matches the true count of 14. E is the honest near-miss: k = 36 really does give 14, one short. The other slip is at k = 35: dropping the endpoint 700 = 35 x 20 because the range 'ends' there gives 14 and wrongly rejects D.
Q30 · NUM-FAM-060 — Answer: C
Step 1 — 360 = 2^3 x 3^2 x 5. Both a and b must be factors of 360, so write a = 2^x1 x 3^y1 x 5^z1 and b = 2^x2 x 3^y2 x 5^z2. Step 2 — the LCM takes the LARGER exponent on each prime, so the conditions are independent: max(x1, x2) = 3, max(y1, y2) = 2, max(z1, z2) = 1. Step 3 — for a cap e, the ordered exponent pairs with max exactly e number (e+1)^2 - e^2 = 2e + 1: all pairs from 0 to e number (e+1)^2, and those with max at most e-1 number e^2. For 2^3: 2(3) + 1 = 7 For 3^2: 2(2) + 1 = 5 For 5^1: 2(1) + 1 = 3 Step 4 — the primes are independent, so multiply: 7 x 5 x 3 = 105. Check the smallest piece by hand: for max(z1, z2) = 1 the pairs are (0,1), (1,0) and (1,1), three of them, matching 2(1) + 1. Answer: C.
Trap. Answering 576 (option E) by taking all ordered pairs of factors of 360, 24 x 24, without imposing the max condition - most of those pairs have an LCM smaller than 360. Answering 75 (option B) uses 5 x 5 x 3, applying the cap 2 to the prime 2 whose exponent is actually 3; answering 63 (option A) uses 7 x 3 x 3, applying cap 1 to the prime 3 whose exponent is 2. Answering 210 (option D) doubles 105 'for order' when 7 x 5 x 3 already counts ordered pairs.