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Admissions tests / TMUA / Paper 2 / Mathematical proof

Demanding. 12 questions, 12 marks, about 50 minutes.

TMUA Paper 2: Mathematical proof, set 3

Direct deductive proof, proof by cases, proof by contradiction, disproof by counterexample, deducing implications, conjecturing from small cases, and ordering a scrambled proof.

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  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  1. 11 mark

    A student wants to prove: 'If n is an odd integer, then n^2 - 1 is divisible by 8.' Which of the following is a valid direct proof?

    1. A Write n = 2k for some integer k. Then n^2 - 1 = 4k^2 - 1, which is 1 less than a multiple of 4, so n^2 - 1 is divisible by 8.
    2. B Write n = 2k + 1 for some integer k. Then n^2 - 1 = 4k^2 + 2k = 2k(2k + 1), which is divisible by 8 because 2k(2k + 1) is always a multiple of 8.
    3. C Write n = 2k + 1 for some integer k. Then n^2 - 1 = 4k^2 + 4k = 4k(k + 1). Among the two consecutive integers k and k + 1, exactly one is even, so k(k + 1) is even; write k(k + 1) = 2m for some integer m. Then n^2 - 1 = 4(2m) = 8m, which is divisible by 8.
    4. D Write n = 2k + 1 for some integer k. Then n^2 - 1 = 4k^2 + 4k = 4k(k + 1), which is divisible by 4. Since every multiple of 4 is also a multiple of 8, n^2 - 1 is divisible by 8.
  2. 21 mark

    A student wants to prove: 'For every integer n, n(n^2 - 1) is divisible by 6.' Which of the following is a valid direct proof?

    1. A n(n^2 - 1) = n(n-1)(n+1), the product of three consecutive integers. Among any three consecutive integers, at least one is even and exactly one is a multiple of 3, so the product is divisible by both 2 and 3, and hence by 6.
    2. B n(n^2 - 1) = n(n-1)^2, since n^2 - 1 factorises as (n-1)^2. Since (n-1)^2 is always a perfect square, n(n-1)^2 is divisible by 6.
    3. C n(n^2 - 1) = (n-1)n(n+1), the product of three consecutive integers. Among any three consecutive integers, at least one is even, so the product is divisible by 2, and every number divisible by 2 is also divisible by 6.
    4. D n(n^2 - 1) = (n-1)n(n+1), the product of three consecutive integers. Since there are three factors being multiplied together, the product is divisible by 3 factorial, that is, by 3! = 6.
  3. 31 mark

    A student wants to prove: 'For every integer n, n(n+1)(n+2)(n+3) is divisible by 8.' Which of the following is a valid proof by cases?

    1. A Among any four consecutive integers n, n+1, n+2 and n+3, n and n+2 are always the two even numbers, and one of them is a multiple of 4, so the product is divisible by 4 x 2 = 8.
    2. B Among any four consecutive integers, exactly two are even, and these two even numbers give a product divisible by 4 (even x even), so n(n+1)(n+2)(n+3) is divisible by 4, but no further factor of 2 can be guaranteed, so only divisibility by 4 can be concluded.
    3. C Among any four consecutive integers, exactly two are even. Since even numbers are each divisible by 2, and two even numbers multiplied together are always divisible by 8, the product n(n+1)(n+2)(n+3) is divisible by 8.
    4. D Among any four consecutive integers n, n+1, n+2 and n+3, exactly two are even, and these two even numbers are themselves consecutive even integers, so one of the two must be a multiple of 4. The product therefore contains one factor divisible by 4 and a separate factor divisible by 2, so the whole product is divisible by 4 x 2 = 8.
  4. 41 mark

    A student wants to prove: 'If n is an integer that is not divisible by 3, then n^4 leaves remainder 1 when divided by 3.' Which of the following is a valid proof by cases?

    1. A Case n = 3k + 1: n^2 leaves remainder 1 when divided by 3. Case n = 3k + 2: n^2 also leaves remainder 1 when divided by 3. Since n^4 = n^2 + n^2, n^4 leaves remainder 1 + 1 = 2 when divided by 3 in both cases.
    2. B Case n = 3k + 1: n^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1, so n^2 leaves remainder 1 when divided by 3, and n^4 = (n^2)^2 therefore leaves the same remainder as 1^2 = 1, that is, remainder 1. Case n = 3k + 2: n^2 = 9k^2 + 12k + 4 = 3(3k^2 + 4k + 1) + 1, again remainder 1, so n^4 also leaves remainder 1. Since n is not divisible by 3, it falls into one of these two cases, and both give n^4 remainder 1 when divided by 3.
    3. C Case n = 3k + 1: n^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1, so n^2 leaves remainder 1, and n^4 = (n^2)^2 leaves remainder 1^2 = 1 when divided by 3. Since this covers the case where n is not divisible by 3, n^4 always leaves remainder 1 when divided by 3.
    4. D Case n = 3k: n^4 = 81k^4, which leaves remainder 1 when divided by 3, matching the other two cases; so the restriction that n is not divisible by 3 is not actually needed, and n^4 leaves remainder 1 when divided by 3 for every integer n.
  5. 51 mark

    In a proof by contradiction that there is no smallest positive rational number, which is the correct starting assumption?

    1. A Assume that there is no smallest positive rational number, and derive a contradiction from that.
    2. B Assume that the smallest positive rational number is 0, and show this is impossible.
    3. C Assume that every rational number is positive, and show this leads to a contradiction.
    4. D Assume that a smallest positive rational number exists, call it q, and show that this leads to a contradiction.
  6. 61 mark

    A student wants to prove: 'If n is an integer and 7n + 4 is odd, then n is odd.' Which of the following is a valid proof by contradiction?

    1. A Assume, for contradiction, that 7n + 4 is odd but n is even. Since n is even, write n = 2k for integer k. Then 7n + 4 = 14k + 4 = 2(7k + 2), which is even. This contradicts the assumption that 7n + 4 is odd, so n must be odd.
    2. B Assume, for contradiction, that 7n + 4 is even and n is even. Since n is even, write n = 2k, so 7n + 4 = 14k + 4 = 2(7k+2), which is even, exactly as assumed, so there is no contradiction and the statement cannot be proved this way.
    3. C Assume, for contradiction, that 7n + 4 is odd but n is even. Since n is even, write n = 2k, so 7n + 4 = 14k + 11, which is odd, exactly as assumed, so there is no contradiction and the statement cannot be proved this way.
    4. D Assume, for contradiction, that n is even. Since n = 2k, 7n + 4 = 14k + 4 = 2(7k+2), which is even. This is what we wanted to show, so the proof is complete.
  7. 71 mark

    A student claims: 'For every positive integer n, n! + 1 is prime.' (Here n! denotes n factorial, so n! = n x (n-1) x ... x 2 x 1.) Which of the following correctly disproves this claim?

    1. A n = 1: 1! + 1 = 2, which is prime, so n = 1 disproves the claim.
    2. B n = 2: 2! + 1 = 3, which is odd, so it is not prime, so n = 2 disproves the claim.
    3. C n = 4: 4! + 1 = 24 + 1 = 25 = 5 x 5, which is not prime, so n = 4 disproves the claim.
    4. D n = 3: 3! + 1 = 7, which is greater than 5, so it must be composite, so n = 3 disproves the claim.
  8. 81 mark

    A student claims: 'If a and b are integers with a^2 = b^2, then a = b.' Which of the following correctly disproves this claim?

    1. A a = 2, b = 2: a^2 = 4 and b^2 = 4, so a^2 = b^2, and a = b, consistent with the claim, so this disproves the claim.
    2. B a = 3, b = -3: a^2 = 9 and b^2 = 9, so a^2 = b^2, and since both squares are equal, a = b, so this disproves the claim.
    3. C a = 2, b = 3: a^2 = 4 and b^2 = 9, so a^2 is not equal to b^2, and a is not equal to b, so this disproves the claim.
    4. D a = -3, b = 3: a^2 = 9 and b^2 = 9, so a^2 = b^2, but a = -3 is not equal to b = 3, so this disproves the claim.
  9. 91 mark

    You are told: 'If a number is a multiple of 12, then it is a multiple of 4' and 'If a number is a multiple of 4, then it is even.' Given that N is a multiple of 12, which of the following can be validly deduced?

    1. A N is a multiple of 4, but nothing further can be said about whether N is even.
    2. B N is even.
    3. C N is even, and it therefore follows that N must be a multiple of 12.
    4. D N is a multiple of 4, and every multiple of 4 is a multiple of 12.
  10. 101 mark

    By evaluating 1, 1+2, 1+2+4, 1+2+4+8 (giving 1, 3, 7, 15), a student conjectures a formula for the sum of the first n powers of 2, starting from 2^0. Which of the following gives both the correct conjecture and a valid justification for every n, not just the four cases checked?

    1. A Conjecture: the sum of the first n powers of 2, from 2^0 to 2^(n-1), is 2^n - 1. Justification: let S = 2^0 + 2^1 + ... + 2^(n-1). Then 2S = 2^1 + 2^2 + ... + 2^n. Subtracting, 2S - S = 2^n - 2^0, so S = 2^n - 1, which matches the observed values and proves the conjecture for every n.
    2. B Conjecture: the sum is 2^n. Justification: the values 1, 3, 7, 15 are each 1 less than 2, 4, 8, 16, so for large n the sum is essentially 2^n, and since the pattern holds for these four cases, it must hold for every n.
    3. C Conjecture: the sum is n(n+1). Justification: substituting n = 1, 2, 3, 4 into n(n+1) gives 2, 6, 12, 20, which is close to the observed sums 1, 3, 7, 15.
    4. D Conjecture: the sum of the first n powers of 2 is 2^n - 1. Justification: the values 1, 3, 7, 15 match 2^n - 1 for n = 1, 2, 3, 4, and since the pattern holds for these four cases, it must hold for every n.
  11. 111 mark

    A student is asked to prove: 'If n leaves remainder 2 when divided by 5, then n^2 leaves remainder 4 when divided by 5.' The proof has been written as four separate lines, given here in a scrambled order:

    W: So n^2 = 25k^2 + 20k + 4 = 5(5k^2 + 4k) + 4.

    X: Suppose n leaves remainder 2 when divided by 5, so n = 5k + 2 for some integer k.

    Y: Since 5k^2 + 4k is an integer, n^2 is 4 more than a multiple of 5, so n^2 leaves remainder 4 when divided by 5.

    Z: Therefore, if n leaves remainder 2 when divided by 5, then n^2 leaves remainder 4 when divided by 5, as required.

    Which order of these four lines gives a valid direct proof?

    1. A W, X, Y, Z
    2. B X, Y, W, Z
    3. C X, W, Y, Z
    4. D X, W, Z, Y
  12. 121 mark

    Suppose n is a positive integer such that both n and n^2 + 2 are prime. Which of the following must be true?

    1. A n must be even.
    2. B n must be a multiple of 3.
    3. C n must be greater than 10.
    4. D n must leave remainder 1 when divided by 3.

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

  1. Question 1Answer: C

    1. To prove a statement about every odd integer n, write n in the general form n = 2k + 1, where k is an integer.
    2. Then n^2 - 1 = (2k+1)^2 - 1 = 4k^2 + 4k + 1 - 1 = 4k^2 + 4k = 4k(k + 1).
    3. Among any two consecutive integers k and k + 1, exactly one is even, so their product k(k + 1) is even: write k(k + 1) = 2m for an integer m.
    4. Then n^2 - 1 = 4(2m) = 8m, which is exactly 8 times an integer, so n^2 - 1 is divisible by 8, and the answer is C.
    • Why not A: Uses n = 2k, the algebraic form of an even integer, to represent n, which the statement says is odd; the case actually constructed is not an odd number at all, so nothing here bears on the claim.
    • Why not B: Expands (2k + 1)^2 - 1 as 4k^2 + 2k instead of the correct 4k^2 + 4k, an arithmetic slip that drops a factor from the middle term and produces an expression that is not, in general, a multiple of 8 (for k = 1 it gives 6, which is not divisible by 8).
    • Why not D: Asserts that every multiple of 4 is also a multiple of 8, which is false in general (4 itself is a multiple of 4 but not of 8); reaching '4k(k+1) is divisible by 4' is correct, but the extra factor of 2 needed for divisibility by 8 has not been justified.

  2. Question 2Answer: A

    1. n(n^2 - 1) factorises using the difference of two squares as n(n-1)(n+1), the product of the three consecutive integers n-1, n and n+1.
    2. Among any three consecutive integers, at least one is a multiple of 2 (since even and odd numbers alternate), so the product is divisible by 2.
    3. Among any three consecutive integers, exactly one is a multiple of 3 (since remainders on division by 3 cycle through 0, 1, 2), so the product is also divisible by 3.
    4. A number divisible by both 2 and 3 is divisible by 6, since 2 and 3 share no common factor, so n(n^2 - 1) is divisible by 6 for every integer n, matching option A.
    • Why not B: Factorises n^2 - 1 as (n-1)^2 rather than the correct (n-1)(n+1), the difference of two squares, so the expression analysed is not equal to n(n^2 - 1) at all; the further claim that a perfect square is automatically divisible by 6 is also false in general (4 is a perfect square but not divisible by 6).
    • Why not C: Correctly shows the product is divisible by 2 but then asserts that every multiple of 2 is a multiple of 6, which is false in general (2 itself is not divisible by 6); establishing divisibility by 2 alone never gives divisibility by 6.
    • Why not D: Reasons that having three factors makes the product divisible by 3! = 6 purely because three numbers are multiplied, but this is not a valid rule (1 x 3 x 5 also has three factors and is not divisible by 6); the real reason is that one of the three consecutive integers must be even and one must be a multiple of 3, not merely that three numbers are multiplied.

  3. Question 3Answer: D

    1. Every four consecutive integers contain exactly two even numbers, and because the integers are consecutive, these two even numbers are themselves consecutive even integers, differing by 2.
    2. Among any two consecutive even integers (such as 4 and 6, or 8 and 10), one is always a multiple of 4, since multiples of 4 occur every other even number.
    3. So the product n(n+1)(n+2)(n+3) contains one factor that is a multiple of 4 and a separate factor that is a multiple of 2, contributing at least 4 x 2 = 8 to the product.
    4. Hence n(n+1)(n+2)(n+3) is divisible by 8 for every integer n, matching option D.
    • Why not A: Assumes n and n+2 are always the even numbers among the four, but this depends on the parity of n: when n is odd, it is n+1 and n+3 that are even, not n and n+2, so the argument does not cover every case.
    • Why not B: Correctly identifies that the product is divisible by 4, but stops short of noticing that one of the two consecutive even numbers among the four must itself be a multiple of 4; the extra factor of 2 this gives is exactly what is needed to reach divisibility by 8.
    • Why not C: Asserts that any two even numbers multiplied together are automatically divisible by 8, which is false in general (2 x 2 = 4, and 2 x 6 = 12, neither divisible by 8); the argument needs the further fact that one of the two even numbers is itself a multiple of 4, not just that both are even.

  4. Question 4Answer: B

    1. Every integer not divisible by 3 leaves remainder 1 or remainder 2 when divided by 3, so these are the two cases to check.
    2. Case n = 3k + 1: n^2 = (3k+1)^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1, so n^2 leaves remainder 1 when divided by 3, and squaring a number that leaves remainder 1 leaves remainder 1^2 = 1 again, so n^4 leaves remainder 1.
    3. Case n = 3k + 2: n^2 = (3k+2)^2 = 9k^2 + 12k + 4 = 3(3k^2 + 4k + 1) + 1, again remainder 1, so n^4 also leaves remainder 1 by the same squaring argument.
    4. Both residue classes that make up 'not divisible by 3' give n^4 remainder 1 when divided by 3, so the answer is B.
    • Why not A: Treats n^4 as n^2 + n^2 rather than n^2 multiplied by itself, so it adds the remainders (1 + 1 = 2) instead of squaring the remainder (1^2 = 1); this gives the wrong remainder in both cases.
    • Why not C: Only examines the case n = 3k + 1 and never examines n = 3k + 2; since 'n is not divisible by 3' covers both of these residue classes, a proof by cases must dispose of both, and the second is missing here.
    • Why not D: Claims 81k^4 leaves remainder 1 when divided by 3, but 81k^4 = 3(27k^4) exactly, leaving remainder 0, not 1; a multiple of 3 raised to any power is still a multiple of 3, so this case actually shows why the restriction 'n is not divisible by 3' is essential, not why it can be dropped.

  5. Question 5Answer: D

    1. A proof by contradiction begins by assuming the negation of the statement to be proved.
    2. The statement here is 'there is no smallest positive rational number'; its negation is 'there is a smallest positive rational number', so assume such a number exists and call it q.
    3. From this assumption, the number q/2 is also a positive rational number, and q/2 is less than q, contradicting the assumption that q was the smallest positive rational.
    4. The correct starting assumption is therefore option D.
    • Why not A: Assumes the very statement to be proved, that no smallest positive rational number exists, rather than its negation; a proof by contradiction must start from the opposite of the target statement, namely that a smallest positive rational does exist.
    • Why not B: Considers only whether 0 could be the smallest positive rational, but 0 is not positive, so this never engages with an actual candidate for 'the smallest positive rational number'; the negation to test is that some positive rational is smallest, not specifically 0.
    • Why not C: Is not the negation of 'there is no smallest positive rational number'; whether every rational number is positive is an entirely different, and false, claim, since negative rationals such as -1 exist, and refuting it says nothing about smallest elements.

  6. Question 6Answer: A

    1. To prove 'if 7n+4 is odd then n is odd' by contradiction, keep the hypothesis (7n+4 is odd) and assume the negation of the conclusion: n is even.
    2. Since n is even, write n = 2k for an integer k.
    3. Then 7n + 4 = 7(2k) + 4 = 14k + 4 = 2(7k + 2), which is even.
    4. This contradicts the assumption that 7n + 4 is odd, so the assumption 'n is even' must be false, and n is odd, matching option A.
    • Why not B: Negates both parts of the given conditional rather than only the conclusion: a contradiction proof of 'if 7n+4 is odd then n is odd' must keep the hypothesis (7n+4 is odd) and assume only the negation of the conclusion (n is even); assuming 7n+4 is even instead tests a different, unrelated statement.
    • Why not C: Miscalculates 7(2k) + 4 as 14k + 11 instead of the correct 14k + 4, an arithmetic slip; the true computation shows 7n+4 is even whenever n is even, which is exactly the contradiction needed, so the slip manufactures a false 'no contradiction' where a real one exists.
    • Why not D: Never uses the given hypothesis that 7n + 4 is odd, so the assumption actually tested is unrelated to the statement; it then treats 'showing 7n+4 is even' as the goal itself, rather than recognising it as the contradiction that disproves the assumption 'n is even'.

  7. Question 7Answer: C

    1. To disprove a universal claim, 'for every positive integer n, n! + 1 is prime', it is enough to exhibit one positive integer n for which n! + 1 is not prime.
    2. Testing n = 4: 4! = 4 x 3 x 2 x 1 = 24, so 4! + 1 = 25.
    3. 25 = 5 x 5, so 25 is composite, not prime.
    4. n = 4 is a positive integer for which the expression fails to be prime, a valid counterexample, so the answer is C.
    • Why not A: n = 1 gives 2, which is prime; a counterexample to 'the expression is always prime' must be a value where the expression is NOT prime, so a value that keeps it prime supports the claim rather than disproving it.
    • Why not B: Argues 3 is not prime because it is odd, but being odd does not make a number composite, since most primes, including 3 itself, are odd; 3 is in fact prime, so this does not disprove the claim at all.
    • Why not D: Argues 7 is composite because it exceeds 5, but there is no such rule: many primes are greater than 5, including 7 itself, which is prime; this pair does not disprove the claim.

  8. Question 8Answer: D

    1. To disprove 'if a^2 = b^2 then a = b', it suffices to find one pair of integers a and b with a^2 = b^2 but a not equal to b.
    2. Let a = -3 and b = 3. Then a^2 = 9 and b^2 = 9, so a^2 = b^2, satisfying the hypothesis.
    3. But a = -3 and b = 3 are not equal.
    4. This pair satisfies the hypothesis while violating the conclusion, giving a valid counterexample, so the answer is D.
    • Why not A: Gives a = b (both equal to 2), which is exactly what the claim predicts when a^2 = b^2; a counterexample must show a^2 = b^2 holding while a = b fails, and this pair does the opposite.
    • Why not B: Correctly computes a^2 = 9 = b^2, but then wrongly concludes a = b; here a = 3 and b = -3 are not equal, so the option's own stated conclusion is false, even though the underlying numbers would have given a valid counterexample if reasoned correctly.
    • Why not C: Uses a pair where a^2 is not equal to b^2 (4 is not 9), so the hypothesis of the claim is not even satisfied; a counterexample must satisfy the hypothesis (a^2 = b^2) while violating the conclusion (a = b), and this pair does neither correctly.

  9. Question 9Answer: B

    1. The two given statements form a chain: multiple of 12 implies multiple of 4, and multiple of 4 implies even.
    2. Since N is a multiple of 12, the first statement gives that N is a multiple of 4.
    3. Applying the second statement to this result, N being a multiple of 4 gives that N is even.
    4. Chaining the two implications together (multiple of 12 implies multiple of 4 implies even) validly deduces that N is even, matching option B.
    • Why not A: Correctly deduces N is a multiple of 4 from the first statement, but then stops, ignoring that the second given statement, 'if a multiple of 4 then even', lets the chain continue one more step to show N is even.
    • Why not C: After correctly deducing N is even, wrongly reasons backwards to conclude N must be a multiple of 12; 'if a multiple of 12 then even' does not mean 'if even then a multiple of 12', and indeed 2 is even without being a multiple of 12.
    • Why not D: Asserts the converse of the first given statement, that every multiple of 4 is a multiple of 12, which is false in general (4 itself is a multiple of 4 but not of 12); 'if A then B' does not imply 'if B then A'.

  10. Question 10Answer: A

    1. Listing the sums 1, 1+2=3, 1+2+4=7, 1+2+4+8=15 suggests the pattern 2^1-1, 2^2-1, 2^3-1, 2^4-1, leading to the conjecture that the sum of the first n powers of 2 (from 2^0 to 2^(n-1)) is 2^n - 1.
    2. To justify this beyond four cases, let S = 2^0 + 2^1 + ... + 2^(n-1), and consider 2S = 2^1 + 2^2 + ... + 2^n.
    3. Subtracting the first sum from the second, every term except the first and last cancels: 2S - S = 2^n - 2^0, so S = 2^n - 1.
    4. This algebraic argument, the standard telescoping trick for a geometric series, proves the conjecture for every positive integer n, not just the four cases checked, matching option A.
    • Why not B: Proposes the formula 2^n, but this does not even match the observed sums (2^1 = 2, not 1; 2^2 = 4, not 3), so the conjecture itself is wrong, and the justification given never establishes a formula that actually equals the sum for every n.
    • Why not C: Proposes n(n+1), the values of which (2, 6, 12, 20) do not match the observed sums (1, 3, 7, 15) at all; describing them as 'close to' the correct values is not the same as matching them, so this conjecture is simply false.
    • Why not D: States the correct conjecture, 2^n - 1, but the justification only checks four small cases and then asserts the pattern 'must hold' for every n with no general argument; agreement on finitely many cases suggests a conjecture but does not by itself prove it for all n.

  11. Question 11Answer: C

    1. A direct proof of an 'if... then...' statement should begin by assuming the hypothesis: line X, 'Suppose n leaves remainder 2 when divided by 5, so n = 5k + 2 for some integer k.'
    2. The next step performs the algebra that follows from this assumption: line W, 'So n^2 = 25k^2 + 20k + 4 = 5(5k^2+4k) + 4.'
    3. Then the reasoning explains why this form gives the desired property: line Y, 'Since 5k^2+4k is an integer, n^2 is 4 more than a multiple of 5, so n^2 leaves remainder 4 when divided by 5.'
    4. Finally the overall conditional conclusion is stated: line Z, giving the order X, W, Y, Z, which is option C.
    • Why not A: Opens with line W, which computes n^2 = 25k^2 + 20k + 4, before line X has introduced the integer k at all; the algebraic step must come after the assumption that defines k, not before it.
    • Why not B: States the conclusion of the working in line Y, that n^2 leaves remainder 4, immediately after the assumption in line X, before the algebraic step in line W that actually justifies it; the working must precede the claim it supports.
    • Why not D: Places the overall conditional conclusion in line Z before line Y, which explains why 5(5k^2+4k)+4 gives remainder 4 when divided by 5; the final conclusion should come last, after every supporting step is in place.

  12. Question 12Answer: B

    1. Every positive integer n leaves remainder 0, 1 or 2 when divided by 3, so check each case.
    2. If n leaves remainder 1: n^2 leaves remainder 1 (mod 3), so n^2+2 leaves remainder 1+2=3, that is, remainder 0 (mod 3), meaning n^2+2 is divisible by 3. Since n^2+2 is prime and divisible by 3, it must equal 3 itself, giving n^2=1, so n=1, but 1 is not prime, contradicting the hypothesis that n is prime. If n leaves remainder 2: n^2 leaves remainder 4, that is, remainder 1 (mod 3), so n^2+2 again leaves remainder 0 (mod 3), and the same contradiction follows.
    3. So n cannot leave remainder 1 or remainder 2 when divided by 3; n must leave remainder 0, that is, n is a multiple of 3.
    4. Checking n=3 confirms this is achievable: 3 is prime, and n^2+2=11 is also prime, so n must be a multiple of 3, matching option B.
    • Why not A: The only even prime is 2, and checking it directly gives n^2+2 = 6, which is not prime, so n = 2 does not even satisfy the given condition; n cannot be even, let alone must be.
    • Why not C: The value that actually satisfies the hypothesis is n = 3 (with n^2 + 2 = 11, both prime), and 3 is not greater than 10, so this claim is false.
    • Why not D: This is exactly the residue class that the argument rules out: if n leaves remainder 1 (or remainder 2) when divided by 3, then n^2 + 2 turns out to be divisible by 3 and greater than 3, so composite, contradicting that n^2+2 is prime; the residue that survives is remainder 0, not remainder 1.

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