Divisibility Rules — Practice Set
Bank code: NUM-DIV · Section: Number Properties · 30 questions — Easy 5 · Medium 8 · Hard 12 · Extreme 5
A divisibility rule is a shortcut for testing whether d divides N without dividing: digit sum for 3 and 9, last two digits for 4, last three for 8, alternating sum for 11, and always split a composite divisor into COPRIME factors before testing it.
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-DIV-006 · MCQ · 50s
The last two digits of the positive integer N are 3 and 6, in that order (so N ends in 36). Which of the following must be true?
(A) N is divisible by 3
(B) N is divisible by 4
(C) N is divisible by 6
(D) N is divisible by 9
(E) N is divisible by 12
Q2 · NUM-DIV-008 · Select all that apply · 55s
Which of the following numbers are divisible by 3? Select all that apply.
(A) 123
(B) 247
(C) 351
(D) 462
(E) 589
Q3 · NUM-DIV-010 · QC · 50s
N = 4,200
Column A: The remainder when N is divided by 6 Column B: The remainder when N is divided by 5
Q4 · NUM-DIV-007 · Numeric Entry · 45s
What is the smallest integer greater than 100 that is divisible by both 2 and 5? Enter your answer as a number.
Numeric entry — write the number.
Q5 · NUM-DIV-011 · MCQ · 55s
Which of the following is NOT divisible by 9?
(A) 729
(B) 918
(C) 1,008
(D) 2,007
(E) 3,457
Level 2 · Medium — 8 questions · ~11 min
two or three steps, one planted trap each
Q6 · NUM-DIV-015 · MCQ · 85s
In the 4-digit number 83d6, d is a single digit (the digits are 8, 3, d, 6 from left to right). For how many values of d is 83d6 divisible by 8?
(A) 1
(B) 2
(C) 3
(D) 5
(E) 10
Q7 · NUM-DIV-018 · MCQ · 75s
Which of the following is divisible by 6 but NOT by 9?
(A) 162
(B) 216
(C) 405
(D) 312
(E) 486
Q8 · NUM-DIV-024 · MCQ · 80s
Which of the following is divisible by 8?
(A) 5,134
(B) 6,316
(C) 7,414
(D) 9,158
(E) 9,648
Q9 · NUM-DIV-032 · MCQ · 85s
Which of the following is divisible by 11?
(A) 4,855
(B) 4,815
(C) 6,161
(D) 8,144
(E) 9,350
Q10 · NUM-DIV-033 · Select all that apply · 85s
The integer n is divisible by both 8 and 9. Which of the following must also divide n? Select all that apply.
(A) 6
(B) 16
(C) 27
(D) 36
(E) 72
Q11 · NUM-DIV-022 · Numeric Entry · 75s
What is the least positive integer that must be added to 2,838 to produce a multiple of 25? Enter your answer as a number.
Numeric entry — write the number.
Q12 · NUM-DIV-017 · Numeric Entry · 85s
What is the largest 3-digit integer that is divisible by both 8 and 9? Enter your answer as a number.
Numeric entry — write the number.
Q13 · NUM-DIV-019 · QC · 85s
p is a positive integer divisible by 12.
Column A: The remainder when p is divided by 8 Column B: 0
Level 3 · Hard — 12 questions · ~20 min
where 162+ is won or lost
Q14 · NUM-DIV-045 · MCQ · 105s
N is a six-digit integer whose digits, reading from left to right, are 2, d, 5, 8, d, 0 - so the single unknown digit d occupies both the ten-thousands place and the tens place. For how many of the ten possible values of d is N divisible by 12?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
Q15 · NUM-DIV-027 · MCQ · 100s
In the 5-digit number 6A,2A4 the same digit A appears in both blank positions (the digits are 6, A, 2, A, 4 from left to right). If 6A,2A4 is divisible by 11, what is A?
(A) 1
(B) 2
(C) 3
(D) 5
(E) 6
Q16 · NUM-DIV-021 · MCQ · 100s
n is a positive integer and 6n is divisible by 8. Which of the following must be true?
(A) n is divisible by 8
(B) n is divisible by 4
(C) n is even but need not be divisible by 4
(D) n is divisible by 3
(E) n is divisible by 24
Q17 · NUM-DIV-051 · MCQ · 100s
N is a 4-digit integer formed by writing a 2-digit integer twice in a row; for example, the 2-digit integer 37 produces N = 3,737. Which of the following must divide N for every such N?
(A) 3
(B) 7
(C) 11
(D) 13
(E) 101
Q18 · NUM-DIV-052 · MCQ · 100s
N is a 3-digit integer whose hundreds digit is greater than its units digit. M is the 3-digit integer formed by writing the digits of N in reverse order. What is the greatest integer that must divide N - M for every such N?
(A) 9
(B) 11
(C) 33
(D) 99
(E) 198
Q19 · NUM-DIV-036 · QC · 105s
N is a positive integer. The sum of the digits of N is 18, and the last two digits of N are 4 and 0, in that order.
Column A: The greatest integer that must divide N Column B: 72
Q20 · NUM-DIV-023 · QC · 100s
a and b are positive integers and a + b is divisible by 9.
Column A: The remainder when a is divided by 9 Column B: The remainder when b is divided by 9
Q21 · NUM-DIV-053 · QC · 95s
N is a 3-digit integer whose digits sum to 12.
Column A: The remainder when N is divided by 9 Column B: 3
Q22 · NUM-DIV-035 · Numeric Entry · 110s
In the 4-digit number 5d8e, d and e are single digits (the digits are 5, d, 8, e from left to right). How many ordered pairs (d, e) make 5d8e divisible by 36? Enter your answer as a number.
Numeric entry — write the number.
Q23 · NUM-DIV-040 · Numeric Entry · 105s
N = 531,660. How many of the nine integers 2, 3, 4, 5, 6, 8, 9, 11 and 25 divide N evenly? Enter your answer as a number.
Numeric entry — write the number.
Q24 · NUM-DIV-044 · Select all that apply · 105s
N is a positive integer and the sum of the digits of N is 45. Which of the following must be true? Select all that apply.
(A) N is divisible by 9
(B) N is divisible by 3
(C) N is divisible by 45
(D) N is odd
(E) N has at least 5 digits
Q25 · NUM-DIV-037 · Select all that apply · 100s
For which of the following values of n is n^3 - n divisible by 12? Select all that apply.
(A) n = 2
(B) n = 3
(C) n = 5
(D) n = 6
(E) n = 8
Level 4 · Extreme — 5 questions · ~10 min
165+ — expect to need the insight, not the grind
Q26 · NUM-DIV-056 · MCQ · 125s
N is the integer written with exactly k sevens and no other digits, so N = 77...7. What is the smallest positive value of k for which N is divisible by 99?
(A) 9
(B) 11
(C) 18
(D) 22
(E) 99
Q27 · NUM-DIV-055 · QC · 125s
n is a positive integer.
Column A: The remainder when 10^n + 1 is divided by 11 Column B: The remainder when 10^n - 1 is divided by 9
Q28 · NUM-DIV-054 · Numeric Entry · 130s
In the 4-digit number 3d4e, d and e are single digits (the digits are 3, d, 4, e from left to right). What is the greatest value of 3d4e that is divisible by 44? Enter your answer as a number.
Numeric entry — write the number.
Q29 · NUM-DIV-048 · Numeric Entry · 125s
N is the least positive integer that is a multiple of 45 and whose digits are all 8s and 0s (every digit of N is either 8 or 0). How many digits does N have? Enter your answer as a number.
Numeric entry — write the number.
Q30 · NUM-DIV-049 · Select all that apply · 120s
n is a positive integer and E = n(n + 1)(n + 2)(n + 3). Which of the following must divide E for every positive integer n? Select all that apply.
(A) 5
(B) 12
(C) 16
(D) 24
(E) 48
Part B — Answer Key
| Q | ID | Level | Type | Answer |
|---|---|---|---|---|
| 1 | NUM-DIV-006 |
easy | MCQ | B |
| 2 | NUM-DIV-008 |
easy | Select all that apply | A, C, D |
| 3 | NUM-DIV-010 |
easy | QC | C |
| 4 | NUM-DIV-007 |
easy | Numeric Entry | 110 |
| 5 | NUM-DIV-011 |
easy | MCQ | E |
| 6 | NUM-DIV-015 |
medium | MCQ | B |
| 7 | NUM-DIV-018 |
medium | MCQ | D |
| 8 | NUM-DIV-024 |
medium | MCQ | E |
| 9 | NUM-DIV-032 |
medium | MCQ | E |
| 10 | NUM-DIV-033 |
medium | Select all that apply | A, D, E |
| 11 | NUM-DIV-022 |
medium | Numeric Entry | 12 |
| 12 | NUM-DIV-017 |
medium | Numeric Entry | 936 |
| 13 | NUM-DIV-019 |
medium | QC | D |
| 14 | NUM-DIV-045 |
hard | MCQ | C |
| 15 | NUM-DIV-027 |
hard | MCQ | E |
| 16 | NUM-DIV-021 |
hard | MCQ | B |
| 17 | NUM-DIV-051 |
hard | MCQ | E |
| 18 | NUM-DIV-052 |
hard | MCQ | D |
| 19 | NUM-DIV-036 |
hard | QC | A |
| 20 | NUM-DIV-023 |
hard | QC | D |
| 21 | NUM-DIV-053 |
hard | QC | C |
| 22 | NUM-DIV-035 |
hard | Numeric Entry | 3 |
| 23 | NUM-DIV-040 |
hard | Numeric Entry | 5 |
| 24 | NUM-DIV-044 |
hard | Select all that apply | A, B, E |
| 25 | NUM-DIV-037 |
hard | Select all that apply | B, C, E |
| 26 | NUM-DIV-056 |
extreme hard | MCQ | C |
| 27 | NUM-DIV-055 |
extreme hard | QC | D |
| 28 | NUM-DIV-054 |
extreme hard | Numeric Entry | 3740 |
| 29 | NUM-DIV-048 |
extreme hard | Numeric Entry | 10 |
| 30 | NUM-DIV-049 |
extreme hard | Select all that apply | B, D |
Part C — Worked Solutions
Q1 · NUM-DIV-006 — Answer: B
Step 1 — ask which rules depend only on the ending. The test for 4 looks at the last TWO digits and nothing else, so it is decided by the ending alone. Step 2 — apply it: the last two digits form 36, and 36/4 = 9, so 4 divides N. B is forced. Step 3 — the rules for 3, 6, 9 and 12 all need the digit sum, which depends on digits we were never given. Counterexample for A, C, D and E: N = 136. Digit sum = 1 + 3 + 6 = 10, not a multiple of 3, so 136 is divisible by neither 3 nor 6 nor 9 nor 12. But 136/4 = 34. Check: 236/4 = 59, 336/4 = 84, 436/4 = 109. Every number ending in 36 is a multiple of 4. Answer: B.
Trap. Seeing that 36 itself is divisible by 3, 6, 9 and 12 and transferring all of that to N. Only the 4-test reads the last two digits; the 3-, 6-, 9- and 12-tests read the digit sum of the WHOLE number. That mistake picks A, C, D or E, and N = 136 kills all four at once.
Q2 · NUM-DIV-008 — Answer: A, C, D
Use the digit-sum test: a number is divisible by 3 exactly when its digit sum is divisible by 3. A. 123: 1 + 2 + 3 = 6, and 6/3 = 2. TRUE (123 = 3 x 41). B. 247: 2 + 4 + 7 = 13, and 13 is not a multiple of 3. FALSE. C. 351: 3 + 5 + 1 = 9, and 9/3 = 3. TRUE (351 = 3 x 117). D. 462: 4 + 6 + 2 = 12, and 12/3 = 4. TRUE (462 = 3 x 154). E. 589: 5 + 8 + 9 = 22, and 22 is not a multiple of 3. FALSE. Answer: A, C and D.
Trap. Mis-adding 2 + 4 + 7 as 12 instead of 13 and selecting B, which is the single most common way this question goes wrong. The other live error is rejecting 351 because it is odd: divisibility by 3 has nothing to do with parity, and 351 = 3 x 117.
Q3 · NUM-DIV-010 — Answer: C
Column A: test 6 as 2 and 3, which are coprime. 4,200 ends in 0, so it is even: divisible by 2. Digit sum = 4 + 2 + 0 + 0 = 6, a multiple of 3: divisible by 3. So 6 divides 4,200 and the remainder is 0. (4,200/6 = 700.) Column B: 4,200 ends in 0, so 5 divides it and the remainder is 0. (4,200/5 = 840.) Both remainders are 0, so the columns are equal. Check: 4,200 = 6 x 700 exactly and 4,200 = 5 x 840 exactly, so both remainders are 0. Answer: C.
Trap. Reading 'remainder' as 'quotient' and comparing 700 against 840, which gives B. The columns ask for what is LEFT OVER, and both divisions come out exact, so both columns are 0.
Q4 · NUM-DIV-007 — Answer: 110
Step 1 — 2 and 5 are coprime, so a number divisible by both is divisible by 2 x 5 = 10. Step 2 — multiples of 10 are exactly the integers ending in 0. Step 3 — the multiples of 10 near 100 are 100 and 110. The question says GREATER than 100, so 100 is excluded. Check: 110/2 = 55 and 110/5 = 22, both integers, and there is no multiple of 10 strictly between 100 and 110. Answer: 110.
Trap. Answering 100, which is divisible by 2 and by 5 but is not greater than 100. The second live error is answering 102: it is even and bigger than 100, but 102 does not end in 0 or 5, so 5 does not divide it.
Q5 · NUM-DIV-011 — Answer: E
Use the digit-sum test for 9: the number is divisible by 9 exactly when its digit sum is. A. 729: 7 + 2 + 9 = 18, a multiple of 9. Divisible. B. 918: 9 + 1 + 8 = 18. Divisible. C. 1,008: 1 + 0 + 0 + 8 = 9. Divisible. D. 2,007: 2 + 0 + 0 + 7 = 9. Divisible. E. 3,457: 3 + 4 + 5 + 7 = 19, and 19 is not a multiple of 9 (18 is, 27 is). NOT divisible. Check: 3,457/9 = 384 remainder 1, matching 19 leaving remainder 1 on division by 9. Answer: E.
Trap. Using the divisibility-by-3 test and stopping there. Every digit sum here except 19 is a multiple of 3 as well as 9, so the 3-test alone does not separate them; and a student who checks only that 1,008 ends in 8 (not 9) wrongly picks C. The 9-test needs the digit sum to be a multiple of 9, not merely of 3.
Q6 · NUM-DIV-015 — Answer: B
Step 1 — the test for 8 reads the last THREE digits, which here are 3d6, that is the number 300 + 10d + 6 = 306 + 10d. Step 2 — run d from 0 to 9: d = 0: 306 = 8 x 38 + 2. No. d = 1: 316 = 8 x 39 + 4. No. d = 2: 326 = 8 x 40 + 6. No. d = 3: 336 = 8 x 42. Works. d = 4: 346 = 8 x 43 + 2. No. d = 5: 356 = 8 x 44 + 4. No. d = 6: 366 = 8 x 45 + 6. No. d = 7: 376 = 8 x 47. Works. d = 8: 386 = 8 x 48 + 2. No. d = 9: 396 = 8 x 49 + 4. No. Step 3 — the working values are d = 3 and d = 7: two of them. Faster: 306 leaves remainder 2 on division by 8 and 10d leaves remainder 2d, so we need 2 + 2d to be a multiple of 8, i.e. d + 1 a multiple of 4, i.e. d = 3 or d = 7. Check: 8,336/8 = 1,042 and 8,376/8 = 1,047. Answer: B. Two values.
Trap. Confusing the 8-rule with the 4-rule and reading only the last TWO digits, d6: that is a multiple of 4 for every odd d, which reports five values and is choice D. The digit-sum errors are the other family: 8 + 3 + d + 6 = 17 + d treated as a multiple of 9 forces d = 1 and gives 1, choice A, while treated as a multiple of 3 it allows d = 1, 4, 7 and gives 3, choice C. Digit sums settle 3 and 9 only. Choice E is for stopping at the 2-test, since the last digit 6 is even whatever d is.
Q7 · NUM-DIV-018 — Answer: D
Two conditions: even AND digit sum a multiple of 3 (that is divisibility by 6), but digit sum NOT a multiple of 9. A. 162: even, digit sum 9. Divisible by 6 and also by 9. Rejected. B. 216: even, digit sum 9. Divisible by 9. Rejected. C. 405: digit sum 9, so divisible by 9 - but it ends in 5, so it is odd and not divisible by 6. Rejected on both counts. D. 312: even, and digit sum 3 + 1 + 2 = 6, a multiple of 3 but NOT of 9. Divisible by 6, not by 9. This is the one. E. 486: even, digit sum 18. Divisible by 9. Rejected. Check: 312/6 = 52 exactly, and 312/9 = 34.67, not an integer. Answer: D. 312.
Trap. Choice C is the option that catches a student who runs only the digit-sum half of the 6-test: 405 has digit sum 9, so it passes for 3, but 405 is odd and 6 needs an even number. Choices A, B and E catch the opposite slip - reading a digit sum of 9 as the signature of 6 and ignoring the 'NOT by 9' condition, when digit sum 9 is precisely what makes a number divisible by 9.
Q8 · NUM-DIV-024 — Answer: E
The test for 8 reads the last THREE digits. A. 5,134: 134/8 = 16.75. No. B. 6,316: 316/8 = 39.5. No. C. 7,414: 414/8 = 51.75. No. D. 9,158: 158/8 = 19.75. No. E. 9,648: 648/8 = 81 exactly. Yes. The halving route for E: 9,648 -> 4,824 -> 2,412 -> 1,206. Three clean halvings, so 8 divides it. Check: 8 x 1,206 = 9,648. Answer: E. 9,648.
Trap. Choice B is built for the student who stops at the last TWO digits: 16 is a multiple of 4, so 6,316 passes the 4-test and fails the 8-test. Choice C is built for the digit-sum error - 7 + 4 + 1 + 4 = 16 is a multiple of 8, but digit sums test only 3 and 9. Choice D is for reading 'ends in 8' as divisible by 8, and choice A for 'it is even, so 8 divides it'.
Q9 · NUM-DIV-032 — Answer: E
Use the alternating digit sum, taking positions from the right: (1st + 3rd) - (2nd + 4th). The number is divisible by 11 exactly when that result is 0 or a multiple of 11. A. 4,855: (5 + 8) - (5 + 4) = 13 - 9 = 4. No. B. 4,815: (5 + 8) - (1 + 4) = 13 - 5 = 8. No. C. 6,161: (1 + 1) - (6 + 6) = 2 - 12 = -10. No. D. 8,144: (4 + 1) - (4 + 8) = 5 - 12 = -7. No. E. 9,350: (0 + 3) - (5 + 9) = 3 - 14 = -11, a multiple of 11. Yes. Check: 9,350/11 = 850 exactly. Answer: E. 9,350.
Trap. Choice A is the 'digit sum is a multiple of 11' error: 4 + 8 + 5 + 5 = 22, which proves nothing, since plain digit sums test only 3 and 9. Choice B is the same error aimed at 9 (digit sum 18). Choice C is for splitting the number in half and subtracting, 61 - 61 = 0, which is the grouping trick for 7 and 13, not the rule for 11. Choice D is for applying the last-two-digits idea: 44 is a multiple of 11, but that rule belongs to 4. Note too that E's alternating sum came out negative; -11 is still a multiple of 11, so do not discard it for the sign.
Q10 · NUM-DIV-033 — Answer: A, D, E
Step 1 - 8 = 2^3 and 9 = 3^2 share no prime factor, so they are coprime. When two coprime integers both divide n, their product divides n: 8 x 9 = 72 divides n. Write n = 72k. Step 2 - an integer is forced to divide n for every such n exactly when it divides 72. Anything that fails on n = 72 itself is not forced. A. 6: 72 = 6 x 12, so n = 6 x 12k. TRUE. B. 16: 16 = 2^4 but 72 = 2^3 x 3^2 carries only three factors of 2. Counterexample n = 72: 72/16 = 4.5. FALSE. C. 27: 27 = 3^3 but 72 carries only two factors of 3. Counterexample n = 72: 72/27 is not an integer. FALSE. D. 36: 72 = 36 x 2, so n = 36 x 2k. TRUE. E. 72: forced in Step 1. TRUE. Nothing here needs n to be positive. n = -72 is divisible by 8 and by 9, and 6, 36 and 72 all divide it; divisibility asks only that the remainder be 0, and sign does not affect that. n = 0 is divisible by every non-zero integer, so it obeys the same list. Check: n = 216 = 72 x 3 is divisible by 8 (last three digits 216 = 8 x 27) and by 9 (digit sum 9). 216/6 = 36, 216/36 = 6, 216/72 = 3, while 216/16 = 13.5 and 216/27 = 8. So 27 divides this particular n even though it is not forced - which is why the test case has to be 72 itself. Answer: A, D and E.
Trap. Over-reaching on one prime at a time. Choice B comes from 'n is divisible by 8 and n is even, so 16 divides n' - n = 72 is divisible by 8 and 72/16 = 4.5. Choice C is the same slip on the 3 side: the digit sum passes the 9-test, so one more factor of 3 gets demanded and 27 is selected, but 72/27 is not an integer. The coprime rule gives 8 x 9 = 72 and stops there.
Q11 · NUM-DIV-022 — Answer: 12
Step 1 — the 25-test reads the last two digits: an integer is divisible by 25 exactly when its last two digits are 00, 25, 50 or 75. Step 2 — the last two digits of 2,838 are 38. Step 3 — moving upward, the next allowed ending is 50, and getting from 38 to 50 costs 50 - 38 = 12. That addition does not disturb the thousands or hundreds digits, so the result is 2,850. Step 4 — nothing smaller works: 2,839 through 2,849 end in 39 through 49, and none of those is 00, 25, 50 or 75. Check: 2,850/25 = 114. Answer: 12.
Trap. Running the 5-test instead of the 25-test: the next integer ending in 0 or 5 is 2,840, which suggests the answer 2. But 2,840/25 = 113.6 - every multiple of 25 is a multiple of 5, not the reverse. The other live error is jumping all the way to the next multiple of 100 and answering 2,900 - 2,838 = 62, when 2,850 already ends in 50.
Q12 · NUM-DIV-017 — Answer: 936
Step 1 — 8 and 9 are coprime, so being divisible by both is the same as being divisible by 8 x 9 = 72. Step 2 — find the largest multiple of 72 below 1,000: 72 x 13 = 936, and 72 x 14 = 1,008, which has 4 digits. Step 3 — so the answer is 936. Check with the two rules directly: the last three digits are 936 and 936/8 = 117, so 8 divides it; the digit sum is 9 + 3 + 6 = 18, a multiple of 9, so 9 divides it. Answer: 936.
Trap. Testing only the 9-rule and answering 999, whose digit sum is 27 - but 999 is odd, so 8 cannot divide it. The mirror error is testing only the 8-rule and answering 992 (992/8 = 124), whose digit sum is 20, not a multiple of 9.
Q13 · NUM-DIV-019 — Answer: D
The temptation is to say that 12 contains a factor of 4 and p is even, so 8 must divide p. Test it instead. Case p = 12: 12 = 8 x 1 + 4, so Column A = 4. Column A is greater. Case p = 24: 24 = 8 x 3 + 0, so Column A = 0. The columns are equal. Two different relationships occur, so nothing is determined. Why it splits: p = 12k = 4 x 3k, so p carries exactly two factors of 2 when k is odd and at least three when k is even. Divisibility by 8 needs three factors of 2, which 12 alone does not supply. Check: p = 12 gives 12 = 8 x 1 + 4 (Column A = 4), p = 24 gives 24 = 8 x 3 (Column A = 0). Answer: D.
Trap. Concluding 'p is divisible by 4 and p is even, so p is divisible by 4 x 2 = 8' and keying C. Those two factors are not coprime, so they cannot be multiplied together; p = 12 is divisible by 4 and even yet leaves remainder 4 on division by 8. Note also that a remainder can never be less than 0, so choice B is impossible here for any p.
Q14 · NUM-DIV-045 — Answer: C
Step 1 — split 12 into COPRIME factors: 12 = 3 x 4. Test 3 and 4 separately; both must pass. Step 2 — the 3-test. Digit sum = 2 + d + 5 + 8 + d + 0 = 15 + 2d. Need 15 + 2d divisible by 3. Since 15 is already a multiple of 3, we need 2d divisible by 3, so d must be a multiple of 3: d = 0, 3, 6 or 9. Step 3 — the 4-test. The last two digits are d and 0, forming the number 10d. Need 10d divisible by 4. Checking d = 0 through 9: 0, 20, 40, 60, 80 work and 10, 30, 50, 70, 90 do not. So d must be even. Step 4 — intersect the two lists. Multiples of 3: {0, 3, 6, 9}. Even digits: {0, 2, 4, 6, 8}. Common: d = 0 and d = 6. Step 5 — d = 0 is a legitimate value. The last two digits are then 00, the number 0, and 0 is divisible by every non-zero integer, so the 4-test passes. Discarding d = 0 out of habit loses half the answer. Step 6 — that is 2 values. Check d = 0: 205,800, and 205,800/12 = 17,150. Check d = 6: 265,860, and 265,860/12 = 22,155. Check d = 3: 235,830 fails the 4-test, since its last two digits form 30 and 30/4 is not an integer. Answer: C. 2.
Trap. Splitting 12 as 2 x 6 instead of 3 x 4. Those factors are not coprime, so the pair of tests they produce is not enough: the last digit is 0, so the 2-test passes for every d, and the 6-test only repeats the 3-test, leaving d = 0, 3, 6, 9 and the answer 4 - choice E. Choice B comes from running the 9-test out of habit (15 + 2d = 27 forces d = 6 alone). Choice D comes from keeping the 3-test but discarding d = 0 on the belief that a blank digit cannot be 0, leaving d = 3, 6, 9 and reporting 3. Choice A comes from applying a digit-sum test to 12 itself: 15 + 2d runs from 15 to 33 and never hits a multiple of 12, so that route reports no values at all. Digit sums test 3 and 9 only.
Q15 · NUM-DIV-027 — Answer: E
Step 1 — label the positions from the right: 4 is position 1, A is position 2, 2 is position 3, A is position 4, 6 is position 5. Step 2 — alternating sum = (odd positions) - (even positions) = (4 + 2 + 6) - (A + A) = 12 - 2A. Step 3 — for divisibility by 11, 12 - 2A must be 0 or a multiple of 11. Step 4 — A is a digit, so 2A runs from 0 to 18 and 12 - 2A runs from 12 down to -6. The multiples of 11 in that range are 11 and 0. 12 - 2A = 0 gives A = 6. 12 - 2A = 11 gives 2A = 1, so A = 0.5 - not a digit. Step 5 — A = 6 is the only value. Check: the number is 66,264, and 66,264/11 = 6,024. Answer: E. A = 6.
Trap. Using the plain digit sum, 12 + 2A, in place of the alternating sum: setting it to the multiple of 11 that is 22 gives A = 5, choice D, and setting it to a multiple of 9 gives 12 + 2A = 18 and A = 3, choice C. Digit sums test 3 and 9, never 11. Choice A comes from counting only one of the two A's, (4 + 2 + 6) - A = 11, which gives A = 1. Choice B comes from looking at the two-digit block '2A' and forcing it to be 22, the multiple of 11 that starts with 2.
Q16 · NUM-DIV-021 — Answer: B
Step 1 — write the condition in prime factors. 6n = 2 x 3 x n, and 8 = 2^3. Step 2 — the factor 6 contributes exactly one 2. The 3 in 6 is useless here, since 8 has no factor of 3. Step 3 — so n must supply the remaining two 2s: n must be divisible by 2^2 = 4. Step 4 — check that 4 is enough: if n = 4, then 6n = 24 and 24/8 = 3. It works, so n does NOT have to be 8, 24 or a multiple of 3. Step 5 — check that 4 is necessary: if n = 2, then 6n = 12 and 12/8 = 1.5. Fails. So 'even' alone is not enough, which kills C. Answer: B. n is divisible by 4.
Trap. Cancelling carelessly: reading '6n divisible by 8' as 'n divisible by 8' and choosing A, which n = 4 refutes (6 x 4 = 24 is divisible by 8 while 4 is not). Choice C is the opposite slip - noticing that n must be even and stopping, when n = 2 already fails. Choice D imports the 3 from the 6, but 8 contains no 3 at all. Choice E multiplies both imports together, demanding 8 x 3 = 24, as if n had to supply the 3 as well as the missing 2s; n = 4 satisfies the stem and is not a multiple of 24.
Q17 · NUM-DIV-051 — Answer: E
Step 1 - name the 2-digit integer. Let it be 10a + b, where a runs from 1 to 9 and b from 0 to 9. Step 2 - write N out in place value. N = 1,000a + 100b + 10a + b = 1,010a + 101b = 101(10a + b). Step 3 - the factor 101 is explicit, so 101 divides N for every such N. That settles E as true. Step 4 - test the others against a single example. Take the 2-digit integer 10, so N = 1,010 = 101 x 10. A. 3: the digit sum of 1,010 is 1 + 0 + 1 + 0 = 2, not a multiple of 3. FAILS. B. 7: 1,010/7 = 144.29. FAILS. C. 11: the alternating sum of 1,010 from the right is (0 + 0) - (1 + 1) = -2, not 0 and not a multiple of 11. FAILS. D. 13: 1,010/13 = 77.69. FAILS. Step 5 - 101 is prime, so the factorisation gives nothing beyond 101 and the original 2-digit integer. Only E survives. Check: 3,737 = 101 x 37, 8,585 = 101 x 85, 1,010 = 101 x 10. Every such N is 101 times the 2-digit integer it was built from. Answer: E. 101.
Trap. Importing the 6-digit fact. Writing a THREE-digit integer twice gives abcabc = abc x 1,001, and 1,001 = 7 x 11 x 13 - that is where choices B, C and D come from. The 4-digit version multiplies by 101, not 1,001, and 101 is prime. Choice C is also where a hurried 11-test lands: the alternating sum of abab is (b + b) - (a + a) = 2(b - a), a multiple of 11 only when a = b, as in 1,111. Choice A comes from the digit sum, which is twice the digit sum of the 2-digit integer; doubling does not create a factor of 3, and 1,010 has digit sum 2.
Q18 · NUM-DIV-052 — Answer: D
Step 1 — name the digits. Let N have hundreds digit a, tens digit b, units digit c, so N = 100a + 10b + c and M = 100c + 10b + a. Step 2 — subtract. N - M = (100a + 10b + c) - (100c + 10b + a) = 99a - 99c = 99(a - c). Step 3 — the tens digit cancels completely, so N - M is always 99 times the difference of the outer digits. Step 4 — 99 divides 99(a - c) for every a and c, so 99 always works - and so do its divisors 9, 11 and 33. Step 5 — can anything larger be forced? Take a - c = 1, for example N = 201 and M = 102: N - M = 99. Nothing bigger than 99 divides 99, so 198 fails. Check: N = 731, M = 137, N - M = 594 = 99 x 6. N = 852, M = 258, N - M = 594 again. N = 923, M = 329, N - M = 594. Every difference is a multiple of 99. Answer: D. 99.
Trap. Choosing A or B by finding one rule that works and stopping: 9 divides N - M and 11 divides N - M, but the question asks for the GREATEST forced divisor, and 99 = 9 x 11 beats both (9 and 11 are coprime, so both dividing forces 99 to divide). Choice C is the half-finished version of that multiplication: spotting the factors 3 and 11 inside 99 and multiplying those two to get 33, which drops the second factor of 3. Choice E is for assuming a - c is at least 2, or for doubling 99 after testing only an example such as N = 852 where N - M = 594 happens to be divisible by 198; N = 201 gives N - M = 99 and settles it.
Q19 · NUM-DIV-036 — Answer: A
Step 1 - what the digit sum gives. A digit sum of 18 is a multiple of 9, so 9 divides N. Step 2 - what the last two digits give. The 4-test reads the last TWO digits: 40 = 4 x 10, so 4 divides N. The last digit is 0, so 5 divides N as well. Step 3 - combine correctly. 4 and 5 are coprime, so 20 divides N. 9 and 20 are coprime, so 9 x 20 = 180 divides N. Step 4 - is 180 the greatest forced divisor? Take two numbers that fit the stem: N = 5,940 (digit sum 18, ends in 40) = 180 x 33, and N = 6,840 (digit sum 18, ends in 40) = 180 x 38. 33 and 38 share no common factor, so nothing larger than 180 can divide both. Column A = 180. Step 5 - 180 > 72, so Column A is greater. Check: 5,940/180 = 33 and 6,840/180 = 38, while 5,940/72 = 82.5 - so 72 does not even divide every such N. Answer: A.
Trap. Reading the 8-rule off the two-digit block 40. Since 40 is a multiple of 8, it looks as though 8 divides N, and 8 x 9 = 72 makes the columns equal - choice C. But the 8-test needs the last THREE digits and the hundreds digit is free: 5,940 ends in 940 and 940/8 = 117.5. The other live error is stopping at 9 x 4 = 36 and forgetting that the final 0 also supplies a factor of 5; 36 < 72 gives choice B.
Q20 · NUM-DIV-023 — Answer: D
Step 1 — the 9-rule in its stronger form: an integer leaves the same remainder on division by 9 as its digit sum does, and remainders add. So if a leaves remainder r, then b must leave remainder 9 - r (or 0 when r = 0) for the sum to be a multiple of 9. Step 2 — nothing in the stem pins r down. Divisibility of a SUM does not force either term to be divisible, or even to match the other. Case a = 9, b = 9: a + b = 18, a multiple of 9. Both remainders are 0, so the columns are equal. Case a = 1, b = 8: a + b = 9. Column A = 1 and Column B = 8, so Column B is greater. Case a = 8, b = 1: a + b = 9. Column A = 8 and Column B = 1, so Column A is greater. Step 3 — three different relationships occur, so nothing is determined. Check: 9 divides 1 + 8 = 9 while 9 divides neither 1 nor 8, which is the whole point. Answer: D.
Trap. Concluding that because 9 divides a + b it must divide a and b separately, so both remainders are 0 and the columns are equal - choice C. Divisibility passes INTO a sum, never back out of one: 9 divides 1 + 8 and divides neither term. Choice A or B is the mirror error, reached by testing one pair such as a = 8, b = 1 and never swapping the roles of a and b.
Q21 · NUM-DIV-053 — Answer: C
Step 1 — the fact behind the 9-rule is stronger than the rule itself: every integer leaves the SAME remainder on division by 9 as its digit sum does. Why: 100a + 10b + c = (99a + 9b) + (a + b + c), and 99a + 9b is a multiple of 9, so only a + b + c survives. Step 2 — here the digit sum is 12, and 12 = 9 x 1 + 3, so the remainder is 3. Step 3 — that holds for every 3-digit N with digit sum 12, so Column A = 3 = Column B. Check three different numbers: 903 -> 903 = 9 x 100 + 3, remainder 3. 480 -> 480 = 9 x 53 + 3, remainder 3. 336 -> 336 = 9 x 37 + 3, remainder 3. Same remainder every time. Answer: C.
Trap. Keying D because N is not a single number - 903, 480 and 336 all qualify - without noticing that the digit sum alone fixes the remainder mod 9. The other live error is reading 'digit sum 12 is a multiple of 3' and concluding the remainder is 0, which gives Column B greater, choice B; that argument settles division by 3, not by 9.
Q22 · NUM-DIV-035 — Answer: 3
Step 1 — split 36 into COPRIME factors: 36 = 4 x 9. Both tests must pass. Step 2 — the 4-test reads the last two digits, 8e, that is 80 + e. Since 80 is already a multiple of 4, we need e divisible by 4: e = 0, 4 or 8. Step 3 — the 9-test reads the digit sum: 5 + d + 8 + e = 13 + d + e, which must be a multiple of 9. Step 4 — take the three allowed values of e in turn. e = 0: need 13 + d a multiple of 9. d runs 0 to 9, so 13 + d runs 13 to 22; the only multiple of 9 there is 18, giving d = 5. Pair (5, 0), number 5,580. e = 4: need 17 + d a multiple of 9, running 17 to 26; the only multiple is 18, giving d = 1. Pair (1, 4), number 5,184. e = 8: need 21 + d a multiple of 9, running 21 to 30; the only multiple is 27, giving d = 6. Pair (6, 8), number 5,688. Step 5 — that is 3 pairs. Check: 5,580/36 = 155, 5,184/36 = 144, 5,688/36 = 158. All integers. Answer: 3.
Trap. Splitting 36 as 6 x 6 and testing divisibility by 6 twice. Those factors are not coprime, and the test they give is only divisibility by 6, which admits far more pairs. The other live error is requiring the digit sum to be a multiple of 3 instead of 9, which triples the count. Answering 9 - the number of values of d alone, ignoring the 4-test on e - is the third.
Q23 · NUM-DIV-040 — Answer: 5
Step 1 - 2: the last digit is 0, which is even. YES. Step 2 - 5: the last digit is 0. YES. Step 3 - 3 and 9: digit sum = 5 + 3 + 1 + 6 + 6 + 0 = 21. 21 is a multiple of 3 but not of 9. So 3 YES, 9 NO. Step 4 - 6: N is divisible by 2 and by 3, and 2 and 3 are coprime. YES. Step 5 - 4: the last two digits are 60, and 60 = 4 x 15. YES. Step 6 - 8: the last THREE digits are 660, and 660/8 = 82.5. NO. Step 7 - 25: the last two digits must read 00, 25, 50 or 75. They read 60. NO. Step 8 - 11: alternating sum from the right = (0 + 6 + 3) - (6 + 1 + 5) = 9 - 12 = -3, which is neither 0 nor a multiple of 11. NO. Step 9 - the integers that divide N are 2, 3, 4, 5 and 6: five of them. Check: 531,660 = 2 x 265,830 = 3 x 177,220 = 4 x 132,915 = 5 x 106,332 = 6 x 88,610. The four failures: 531,660/8 = 66,457.5, /9 = 59,073.33, /11 = 48,332.72, /25 = 21,266.4. Answer: 5.
Trap. Letting the 3-test stand in for the 9-test: the digit sum 21 is a multiple of 3, so 9 gets counted as well and the answer becomes 6. Digit sums settle 3 and 9 separately, and 21 is not a multiple of 9. The mirror error is letting the 4-test stand in for the 8-test - 60 is a multiple of 4, so 8 gets counted too and the answer climbs to 7 - when the 8-test reads the last three digits, 660, which 8 does not divide. Reading 25 off the final 0 adds a third false yes and gives 8: every multiple of 25 is a multiple of 5, not the reverse.
Q24 · NUM-DIV-044 — Answer: A, B, E
A. Digit sum 45 = 9 x 5 is a multiple of 9, so 9 divides N. TRUE. B. 45 is also a multiple of 3, so 3 divides N. (It also follows from A, since 3 divides 9.) TRUE. C. 45 = 9 x 5, and 9 and 5 are coprime, so divisibility by 45 needs the 5-test too - the last digit must be 0 or 5. N = 99,999 has digit sum 45 but ends in 9, and 99,999/45 = 2,222.2. FALSE. D. N = 99,990 has digit sum 9 + 9 + 9 + 9 + 0 = 45 and is even. FALSE. E. Each digit is at most 9, so k digits give a digit sum of at most 9k. To reach 45 we need 9k >= 45, i.e. k >= 5. TRUE, and 99,999 shows 5 digits is attainable. Answer: A, B and E.
Trap. Selecting C by reading the digit-sum rule as if it extended to any divisor: digit sums decide 3 and 9 and nothing else, so 45 needs a separate 5-test that 99,999 fails. Selecting D comes from noticing that 45 is odd and transferring that parity to N; the digit sum's parity says nothing about the number's, as 99,990 shows.
Q25 · NUM-DIV-037 — Answer: B, C, E
Step 1 - factor: n^3 - n = n(n^2 - 1) = (n - 1)n(n + 1), the product of three consecutive integers. Step 2 - among three consecutive integers one is a multiple of 3 and at least one is even, so 6 divides the product for every n. Split 12 into COPRIME factors: 12 = 3 x 4. The 3 is already guaranteed, so everything turns on 4. Step 3 - if n is odd, n - 1 and n + 1 are consecutive even numbers and one of them is a multiple of 4, so the product carries 4 automatically. If n is even, n - 1 and n + 1 are both odd and the only even factor is n itself, so 4 divides the product only when 4 divides n. A. n = 2: 1 x 2 x 3 = 6, and 6/12 is not an integer. n is even and 4 does not divide 2. FALSE. B. n = 3: 2 x 3 x 4 = 24 = 12 x 2. TRUE. C. n = 5: 4 x 5 x 6 = 120 = 12 x 10. TRUE. D. n = 6: 5 x 6 x 7 = 210, and 210/12 = 17.5. n is even and 4 does not divide 6. FALSE. E. n = 8: 7 x 8 x 9 = 504 = 12 x 42. n is even, but 4 divides 8, so the factor of 4 arrives anyway. TRUE. Check: the rule is 'every odd n works, and an even n works exactly when 4 divides n'. The even values offered are 2, 6 and 8; 4 divides only 8, and 8 is the only even one selected. Answer: B, C and E.
Trap. Stopping at Step 2 and selecting all five on the correct but insufficient fact that three consecutive integers are always divisible by 6. Twelve needs a second factor of 2: n = 2 gives 6 and n = 6 gives 210, and 12 divides neither. The opposite error is reading Step 3 as 'n must be odd' and selecting only B and C, dropping E; n = 8 supplies the factor of 4 by itself.
Q26 · NUM-DIV-056 — Answer: C
Step 1 — split 99 into coprime factors: 99 = 9 x 11. N must pass both tests. Step 2 — the 9-test. Every digit is 7, so the digit sum is 7k. We need 7k divisible by 9. Since 7 and 9 are coprime, the 7 cannot help, so k itself must be a multiple of 9: k = 9, 18, 27, ... Step 3 — the 11-test. The alternating sum adds the digits in odd positions and subtracts those in even positions. All digits equal 7, so each adjacent pair cancels: if k is even the alternating sum is 0, and if k is odd it is 7 (one unpaired digit). We need 0 or a multiple of 11, and 7 is neither, so k must be EVEN. Step 4 — combine: k must be a multiple of 9 and even, so k must be a multiple of 18. The smallest positive such k is 18. Check k = 18: digit sum = 7 x 18 = 126 = 9 x 14, so 9 divides N. k is even, so the alternating sum is 0 and 11 divides N. Check k = 9: digit sum 63 passes the 9-test, but 9 is odd so the alternating sum is 7 and the 11-test fails - 777,777,777 is not divisible by 99. Answer: C. 18.
Trap. Running only the 9-test, finding k = 9 and choosing A. The 11-test is the half that is easy to skip because the digits are all the same, and it is exactly what rules 9 out: with an odd count of sevens the alternating sum is 7, not 0. Choice B is for reading '99 needs 11' as 'k = 11'; k = 11 is even-handed on neither test (digit sum 77 is not a multiple of 9). Choice D is for taking the smallest even multiple of 11 rather than combining the two conditions, and choice E for assuming k must reach 99 itself.
Q27 · NUM-DIV-055 — Answer: D
Column B first, because it is fixed. 10^n - 1 is the number written as n nines: 9, 99, 999, and so on. Its digit sum is 9n, a multiple of 9, so 9 divides it and Column B = 0 for every n. Column A moves. 10^n + 1 is a 1, then n - 1 zeros, then a 1, for example 101, 1001, 10001. Apply the 11-test - the alternating sum. n = 1: the number is 11. Alternating sum (1) - (1) = 0, so 11 divides it and Column A = 0. n = 2: the number is 101. Alternating sum (1 + 1) - (0) = 2, not a multiple of 11, so Column A = 101 - 99 = 2. n = 3: the number is 1,001. Alternating sum (1 + 0) - (0 + 1) = 0, so Column A = 0. n = 4: the number is 10,001. Alternating sum (1 + 0 + 1) - (0 + 0) = 2, so Column A = 2. The pattern: the two 1s sit in the same position class when n is even (both odd positions), so they add to 2; they sit in opposite classes when n is odd, so they cancel to 0. Now compare. n = 1: Column A = 0 = Column B, the columns are equal. n = 2: Column A = 2 > 0 = Column B, Column A is greater. Two different relationships occur, so nothing is determined. Check: n = 2 gives 101 = 11 x 9 + 2, so Column A = 2, while 10^2 - 1 = 99 = 9 x 11, so Column B = 0. Answer: D.
Trap. Testing n = 1 alone, getting 0 and 0, and keying C. Column B really is invariant, which makes the coincidence at n = 1 look conclusive; Column A flips with the parity of n, so any single test case is worthless. Keying A after testing only n = 2 is the mirror error. Whenever one column depends on n, test an odd n and an even n before answering.
Q28 · NUM-DIV-054 — Answer: 3740
Step 1 — split 44 into COPRIME factors: 44 = 4 x 11. Both tests must pass. Step 2 — the 4-test reads the last two digits, 4e, that is 40 + e. Since 40 is a multiple of 4, we need e divisible by 4: e = 0, 4 or 8. Step 3 — the 11-test. Positions from the right: e is 1st, 4 is 2nd, d is 3rd, 3 is 4th. Alternating sum = (e + d) - (4 + 3) = d + e - 7. This must be 0 or a multiple of 11. With d and e digits, d + e runs from 0 to 18, so d + e - 7 runs from -7 to 11. The multiples of 11 available are 0 and 11, giving d + e = 7 or d + e = 18. Step 4 — combine with the three allowed values of e, and prefer the largest d since d is the more significant digit. d + e = 18 forces d = e = 9, but e = 9 is not one of 0, 4, 8. Rejected. d + e = 7 with e = 0 gives d = 7: the number 3,740. d + e = 7 with e = 4 gives d = 3: the number 3,344. d + e = 7 with e = 8 would give d = -1, impossible. Step 5 — the two valid numbers are 3,740 and 3,344. The greater is 3,740. Check: 3,740/44 = 85, and 3,344/44 = 76. Both exact, and 3,740 > 3,344. Answer: 3740.
Trap. Splitting 44 as 2 x 22 and testing only 'even' plus 'divisible by 22'. Those factors are not coprime, so the pair is not equivalent to 44. The commoner error is finding one valid number and stopping: a student who lands on 3,344 first answers 3344, missing that d = 7, e = 0 also works and is larger. The stem asks for the GREATEST value, so every allowed e must be checked. Answering 2 - the count of valid numbers rather than the number itself - is the third live error.
Q29 · NUM-DIV-048 — Answer: 10
Step 1 - split 45 into COPRIME factors: 45 = 9 x 5. N has to pass both tests. Step 2 - the 5-test. The last digit must be 0 or 5. The digit 5 is not allowed, so N ends in 0. N therefore needs at least one 0. Step 3 - the 9-test. Let N contain k eights. Every other digit is 0, so the digit sum is 8k, and 9 must divide 8k. Since 8 and 9 are coprime, 9 must divide k. Step 4 - k cannot be 0 (a string of zeros is not a positive integer), so the least k is 9. N needs at least nine 8s plus at least one 0: at least 10 digits. Step 5 - ten digits is achievable, and the least such N puts the single 0 in the last place: N = 8,888,888,880. Step 6 - verify N. Digit sum = 8 x 9 = 72, a multiple of 9, so 9 divides N. The last digit is 0, so 5 divides N. 9 and 5 are coprime, so 45 divides N. Check: 8,888,888,880/45 = 197,530,864. N has 10 digits. Answer: 10.
Trap. Running the 9-test on the COUNT of digits rather than on the digit sum, concluding that nine digits are needed and answering 9. The digit sum is 8k, not k. The related error is testing 9 alone and stopping at 888,888,888, whose digit sum 72 is a multiple of 9 - that number fails the 5-test, and the extra 0 it needs is exactly what pushes the count to 10. Splitting 45 as 3 x 15 instead of 9 x 5 loses the same way: 3 and 15 are not coprime, and a digit sum divisible by 3 is not enough.
Q30 · NUM-DIV-049 — Answer: B, D
E is a product of FOUR consecutive integers. Count the prime factors that are guaranteed. Factors of 2: among four consecutive integers there are exactly two even numbers, and they differ by 2, so one of them is a multiple of 4. That gives at least 4 x 2 = 8, i.e. three factors of 2. Factors of 3: among four consecutive integers at least one is a multiple of 3. So E is always divisible by 2^3 x 3 = 24. And n = 1 gives E = 1 x 2 x 3 x 4 = 24 exactly, so 24 is the whole of what is forced: an integer must divide E for every n precisely when it divides 24. A. 5: 5 does not divide 24. n = 1 gives E = 24 and 24/5 = 4.8. FALSE. B. 12: 12 divides 24, so 12 divides E for every n. TRUE. C. 16: 16 = 2^4 needs a fourth factor of 2. n = 1 gives 24 and 24/16 = 1.5. FALSE. D. 24: guaranteed by the count above, and attained at n = 1. TRUE. E. 48: 48 = 2^4 x 3 has the same missing factor of 2. n = 1 gives 24 and 24/48 = 0.5. FALSE. Check across values: n = 2 gives 120 = 24 x 5, n = 3 gives 360 = 24 x 15, n = 5 gives 1,680 = 24 x 70 - all multiples of 12 and of 24. Note that 5 divides E at n = 2, where E = 120, and 16 divides E at n = 5, where E = 1,680; 'must divide' means for every n, and n = 1 kills A, C and E at once. Answer: B and D.
Trap. Selecting E on the rule of thumb that a product of k consecutive integers is divisible by k factorial and then over-reaching past 4! = 24 to 48. The bound 24 is exact, not a floor: n = 1 gives E = 24 itself and nothing larger than 24 divides 24. Choice C is the same over-reach in smaller form, counting the two even factors as 2 x 2 x 4 = 16 instead of 2 x 4 = 8. Choice A assumes four consecutive integers must contain a multiple of 5, which needs five of them; n = 1 gives 24 again.