TMUA Paper 2: Mathematical proof, set 4 (stretch)

• Answer all questions. No calculator.
• Each question has exactly one correct answer.
• This is a stretch set: questions combine more than one specification point and are pitched beyond what the real exam typically asks. Look for the short algebraic route before grinding through cases.

1. 11 mark

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

   1. A Pair n with (n+1) and (n+2) with (n+3) instead: n(n+1) = n^2 + n and (n+2)(n+3) = n^2 + 5n + 6. Multiplying these two quadratics and adding 1 gives (n^2 + 3n + 1)^2, by the same reasoning as pairing n with (n+3) and (n+1) with (n+2).
   2. B Let y = n^2 + 3n. Since n(n+3) = n^2 + 3n = y and (n+1)(n+2) = n^2 + 3n + 2 = y + 2, the product n(n+1)(n+2)(n+3) equals y(y+2) = y^2 + 2y. Adding 1 gives y^2 + 2y + 1 = (y+1)^2 = (n^2 + 3n + 1)^2, which is a perfect square because n^2 + 3n + 1 is an integer whenever n is.
   3. C Checking n = 1 gives 1 x 2 x 3 x 4 + 1 = 25 = 5^2, and n = 2 gives 2 x 3 x 4 x 5 + 1 = 121 = 11^2. Since the claim holds in both cases checked, this is a valid direct proof that it holds for every integer n.
   4. D n(n+1)(n+2)(n+3) + 1 expands to n^4 + 6n^3 + 11n^2 + 6n + 1. Since (n^2 + 3n)^2 = n^4 + 6n^3 + 9n^2, the expression equals (n^2 + 3n)^2 + 1, which is one more than a perfect square and therefore not itself a perfect square, so the claim is false.
2. 21 mark

   A student wants to prove: 'n(n+1)(2n+1) is divisible by 6 for every integer n.' Which of the following is a valid proof by cases?

   1. A Since n and n+1 are consecutive integers, one of them is even, so n(n+1)(2n+1) is divisible by 2. As every even number is also divisible by 6, the product is divisible by 6.
   2. B Consider n mod 3: if n = 3k, n is divisible by 3; if n = 3k+1, then 2n+1 = 6k+3 is divisible by 3; if n = 3k+2, then n+1 = 3k+3 is divisible by 3. Since every case gives a factor divisible by 3, and every multiple of 3 is also a multiple of 6, the product is divisible by 6.
   3. C Since n and n+1 are consecutive integers, one of them is even, so n(n+1)(2n+1) is divisible by 2. Also, taking n = 1 gives 2n+1 = 3, which is divisible by 3, so 2n+1 is divisible by 3 for every integer n. Since the product is divisible by both 2 and 3, it is divisible by 6.
   4. D Since n and n+1 are consecutive integers, one of them is even, so n(n+1)(2n+1) is divisible by 2. For divisibility by 3, consider n mod 3: if n = 3k, then n is divisible by 3; if n = 3k+1, then 2n+1 = 6k+3 = 3(2k+1) is divisible by 3; if n = 3k+2, then n+1 = 3k+3 = 3(k+1) is divisible by 3. In every case one of the three factors n, n+1, 2n+1 is divisible by 3, so the product is divisible by 3. Since the product is divisible by both 2 and 3, and 2 and 3 share no common factor, it is divisible by 6.
3. 31 mark

   A student wants to prove, by contradiction, that there is no rational number x with x^3 = 2. The proof is allowed to use, without further justification, the fact that 'if an integer's cube is even, then the integer itself is even.' Which of the following is a valid proof by contradiction?

   1. A Assume, for contradiction, that x = a/b for integers a and b with no common factor and b not equal to 0, and that x^3 = 2. Then a^3 = 2b^3, so a^3 is even, and by the given fact a is even; write a = 2c. Substituting gives 8c^3 = 2b^3, so b^3 = 4c^3, which is even, so by the given fact b is even too. But then a and b share the factor 2, contradicting that they have no common factor. So no such rational x exists.
   2. B Assume, for contradiction, that x = a/b for integers a and b (b not equal to 0) and that x^3 = 2. Then a^3 = 2b^3, so a is even by the given fact; write a = 2c, giving b^3 = 4c^3, so b is even too by the given fact. Since a and b are both even, this contradicts x^3 = 2, so no such rational x exists.
   3. C Assume, for contradiction, that x = a/b for integers a and b with no common factor and that x^3 = 2. Then a^3 = 2b^3, so a is even by the given fact; write a = 2c. Substituting gives (2c)^3 = 2c^3 = 2b^3, so b^3 = c^3, meaning b = c. Since this does not show b is even, no contradiction is reached, so this method cannot prove the claim.
   4. D Assume, for contradiction, that x = a/b for integers a and b with no common factor and that x^3 = 2. Then a^3 = 2b^3 is even, and since an even integer always has an even cube, a is even; write a = 2c. Substituting gives b^3 = 4c^3, which is even, so again since an even integer always has an even cube, b is even. This contradicts a and b having no common factor, so no such rational x exists.
4. 41 mark

   Having checked n^2 - n + 11 for n = 1, 2, ..., 10 and found every value prime (11, 13, 17, 23, 31, 41, 53, 67, 83, 101), a student conjectures: 'n^2 - n + 11 is prime for every positive integer n.' Which of the following correctly disproves this conjecture?

   1. A n = 10: n^2 - n + 11 = 100 - 10 + 11 = 101, which is not divisible by any of 2, 3, 5 or 7, so it must be composite, and n = 10 disproves the conjecture.
   2. B n = 11: n^2 - n + 11 = 121 - 11 + 11 = 121 - 11 = 110, which is not prime, so n = 11 disproves the conjecture.
   3. C n = 11: n^2 - n + 11 = 11 x 10 + 11 = 121 = 11^2, which is not prime, so n = 11 disproves the conjecture.
   4. D n = 0: n^2 - n + 11 = 0 - 0 + 11 = 11, which is not prime since it equals 11 x 1, so n = 0 disproves the conjecture.
5. 51 mark

   You are given the following three true statements about a positive integer x:

   (1) If x is a multiple of 18, then x is a multiple of 9.

   (2) If x is a multiple of 9, then the digit sum of x is a multiple of 9.

   (3) The digit sum of x is not a multiple of 9.

   Which of the following can be validly deduced from statements (1) to (3)?

   1. A x is not a multiple of 9, and no further conclusion about whether x is a multiple of 18 can be drawn from statements (1) to (3).
   2. B x is a multiple of 18. Since the digit sum is not a multiple of 9, x is not a multiple of 9 by statement (2). But statement (1) only tells us multiples of 18 ARE multiples of 9, not the reverse, so x being a non-multiple of 9 tells us nothing about x and 18; since nothing rules it out, x must be a multiple of 18.
   3. C The digit sum of x is a multiple of 9.
   4. D x is not a multiple of 18.
6. 61 mark

   By evaluating 1^3, 1^3+2^3, 1^3+2^3+3^3, 1^3+2^3+3^3+4^3 (giving 1, 9, 36, 100), a student conjectures a formula for the sum of the first n cubes. 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 cubes is (n(n+1)/2)^2. Justification: the values 1, 9, 36, 100 match (1x2/2)^2, (2x3/2)^2, (3x4/2)^2, (4x5/2)^2, and since the pattern holds for these four cases, it must hold for every n.
   2. B Conjecture: the sum of the first n cubes is (n(n+1)/2)^2. Justification: writing S(n) = (n(n+1)/2)^2, the difference S(n) - S(n-1) = (n(n+1)/2)^2 - ((n-1)n/2)^2 = (n^2/4)[(n+1)^2 - (n-1)^2] = (n^2/4)(4n) = n^3, which is exactly the cube added when going from the sum of the first (n-1) cubes to the sum of the first n cubes. Since S(0) = 0 matches the empty sum and each step adds the correct next cube, S(n) gives the sum of the first n cubes for every n.
   3. C Conjecture: the sum of the first n cubes is n(n+1)(2n+1)/6. Justification: substituting n = 1, 2, 3, 4 gives 1, 5, 14, 30, which the student takes as agreeing with the observed sums 1, 9, 36, 100.
   4. D Conjecture: the sum of the first n cubes is n^3. Justification: since the nth cube itself is n^3, the sum up to the nth term must also equal n^3.
7. 71 mark

   A student is asked to prove: 'For every integer n, (n+1)^3 - n^3 - 1 is divisible by 6.' The proof has been written as five separate lines, given here in a scrambled order:

   V: Since n and n+1 are consecutive integers, one of them must be even, so n(n+1) is even; write n(n+1) = 2m for some integer m.

   W: Let n be any integer. Expanding, (n+1)^3 = n^3 + 3n^2 + 3n + 1.

   X: Subtracting n^3 and 1 from both sides of that equation gives (n+1)^3 - n^3 - 1 = 3n^2 + 3n = 3n(n+1).

   Y: Substituting n(n+1) = 2m into the expression from line X gives (n+1)^3 - n^3 - 1 = 3(2m) = 6m, so the expression is 6 times an integer.

   Z: Therefore, for every integer n, (n+1)^3 - n^3 - 1 is divisible by 6, as required.

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

   1. A W, X, V, Y, Z
   2. B V, W, X, Y, Z
   3. C W, X, Y, V, Z
   4. D X, W, V, Y, Z
8. 81 mark

   Suppose a, b and c are positive integers with a^2 + b^2 = c^2, and c is odd. Which of the following must be true?

   1. A Both a and b are even.
   2. B Both a and b are odd.
   3. C Exactly one of a and b is even, and the other is odd.
   4. D Whether a and b are even or odd depends on the specific values chosen, and no general conclusion can be drawn.
9. 91 mark

   Having checked that 2^2-1=3, 2^3-1=7, 2^5-1=31 and 2^7-1=127 are all prime (with 2, 3, 5, 7 themselves prime), a student claims: 'If n is prime, then 2^n - 1 is prime.' Which of the following correctly disproves this claim?

   1. A n = 9: 2^9 - 1 = 511 = 7 x 73, which is not prime, so n = 9 disproves the claim.
   2. B n = 11: 11 is prime, and 2^11 - 1 = 2047 = 23 x 89, which is not prime, so n = 11 disproves the claim.
   3. C n = 13: 13 is prime, and 2^13 - 1 = 8191, which is even, so it cannot be prime, hence n = 13 disproves the claim.
   4. D n = 11: 11 is prime, and 2^11 - 1 = 2^10 = 1024, which is not prime, so n = 11 disproves the claim.
10. 101 mark

    Let x be a real number. Consider the statements P: 'x^2 - 5x + 6 = 0' and Q: 'x = 2'. Which of the following best describes the logical relationship between P and Q?

    1. A Q is necessary and sufficient for P.
    2. B Q is necessary for P, but not sufficient for P.
    3. C Q is neither necessary nor sufficient for P.
    4. D Q is sufficient for P, but not necessary for P.
11. 111 mark

    A student wants to prove, by contradiction, that log_2(3) is irrational (where log_2 denotes log base 2). Which of the following is a valid proof?

    1. A Assume, for contradiction, that log_2(3) is irrational. Then it cannot be written as a/b for integers a and b, which is exactly the definition of irrational, so the proof is complete.
    2. B Assume, for contradiction, that log_2(3) = a/b for positive integers a and b. Then 2^a = 3^b. Since both a and b are positive integers, 2^a and 3^b are both even, so no contradiction arises, and the method fails to prove the claim.
    3. C Assume, for contradiction, that log_2(3) = a/b for positive integers a and b. Then 2^(a/b) = 3, so raising both sides to the power b gives 2^a = 3^b. But 2^a is even for every positive integer a, while 3^b is odd for every positive integer b (an odd number raised to any power stays odd), so 2^a and 3^b can never be equal. This contradiction shows log_2(3) cannot be written as a/b for positive integers a and b, so it is irrational.
    4. D Assume, for contradiction, that log_2(3) = a/b for positive integers a and b. Then 2^a = 3^b. Trying a = 2, b = 1 gives 2^2 = 4 and 3^1 = 3, which are not equal, so this disproves the assumption and completes the proof.
12. 121 mark

    A student wants to prove: 'For every real number x, x^2 - 6x + 11 > 0.' Which of the following is a valid direct proof?

    1. A Completing the square: x^2 - 6x + 11 = (x-3)^2 - 9 + 11 = (x-3)^2 + 2. Since (x-3)^2 >= 0 for every real number x, (x-3)^2 + 2 >= 2 > 0, so x^2 - 6x + 11 > 0 for every real x.
    2. B Completing the square: x^2 - 6x + 11 = (x-3)^2 + 11. Since (x-3)^2 >= 0 for every real number x, (x-3)^2 + 11 >= 11 > 0, so x^2 - 6x + 11 > 0 for every real x.
    3. C The discriminant of x^2 - 6x + 11 is (-6)^2 - 4(1)(11) = 36 - 44 = -8, which is negative, so the quadratic has no real roots. Since it has no real roots, x^2 - 6x + 11 is never zero, so it must be positive for every real x.
    4. D Substituting x=0 gives 11 > 0, x=3 gives 2 > 0, and x=6 gives 11 > 0. Since the expression is positive at these three values, x^2 - 6x + 11 > 0 for every real number x.

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: B

   1. Pair the first and last factors, and the middle two factors: n(n+3) = n^2 + 3n, and (n+1)(n+2) = n^2 + 3n + 2. Both are expressed in terms of y = n^2 + 3n.
   2. The product is then y(y+2) = y^2 + 2y, and adding 1 gives y^2 + 2y + 1 = (y+1)^2.
   3. Since y = n^2 + 3n is an integer whenever n is, y + 1 = n^2 + 3n + 1 is also an integer, so (y+1)^2 = (n^2+3n+1)^2 is a perfect square.
   4. This single algebraic identity holds for every integer n at once, which is a genuine direct proof, unlike checking a few cases or mis-expanding the square, so the answer is B; this is much faster than expanding the full quartic and trying to spot the square from scratch.
   • Why not A: Pairs n with (n+1) and (n+2) with (n+3) rather than n with (n+3) and (n+1) with (n+2); the two resulting quadratics, n^2+n and n^2+5n+6, differ by 4n+6, not by a constant, so the y(y+2)+1=(y+1)^2 trick does not apply to this pairing, and the final line is simply asserted rather than derived from it.
   • Why not C: Verifies only n=1 and n=2 and then asserts the pattern continues 'for every integer n' with no general argument; agreement on finitely many cases is evidence for a conjecture, not a proof of one.
   • Why not D: Squares n^2+3n instead of n^2+3n+1, so it misses the cross-terms contributed by the '+1' (an extra 2n^2+6n); this arithmetic slip also flips the conclusion, since the correct grouping shows the expression IS a perfect square for every integer n.

2. Question 2Answer: D

   1. Divisibility by 2: n and n+1 are consecutive integers, so exactly one of them is even, making n(n+1) even and hence n(n+1)(2n+1) even.
   2. Divisibility by 3: every integer n leaves remainder 0, 1 or 2 on division by 3. If n = 3k, n itself is divisible by 3. If n = 3k+1, then 2n+1 = 6k+3 = 3(2k+1) is divisible by 3. If n = 3k+2, then n+1 = 3k+3 = 3(k+1) is divisible by 3.
   3. These three cases are exhaustive, and in each one a different one of the three factors n, n+1, 2n+1 is divisible by 3, so the product is always divisible by 3.
   4. Since the product is divisible by both 2 and 3, and gcd(2,3)=1, it is divisible by their product, 6, matching option D; establishing the two prime factors 2 and 3 separately, rather than one and hoping the other follows for free, is what actually proves divisibility by 6.
   • Why not A: Shows the product is divisible by 2, which is correct, but then asserts 'every even number is divisible by 6', which is false (for example 8 is even but not divisible by 6); divisibility by 6 needs the extra factor of 3 established separately, which this option never does.
   • Why not B: Shows the product is divisible by 3, which is correct, but then asserts 'every multiple of 3 is also a multiple of 6', which is false (for example 9 is a multiple of 3 but not of 6); divisibility by 6 needs the extra factor of 2 established separately, which this option never does.
   • Why not C: Treats the single case n=1, where 2n+1=3 is divisible by 3, as if it showed 2n+1 is divisible by 3 for every n; it is not (n=2 gives 2n+1=5), so the argument for the factor of 3 is an unjustified generalisation from one example rather than a case analysis covering every remainder mod 3.

3. Question 3Answer: A

   1. Assume, for contradiction, that x = a/b for integers a and b with no common factor, b not equal to 0, and x^3 = 2.
   2. Then a^3 = 2b^3, so a^3 is even, and by the given fact a itself is even; write a = 2c for an integer c.
   3. Substituting: (2c)^3 = 8c^3 = 2b^3, so b^3 = 4c^3, which is even, so by the given fact b is even too.
   4. But then a and b are both even, sharing the factor 2, which contradicts a and b having no common factor. So the assumption was false, and no rational x with x^3 = 2 exists, matching option A.
   • Why not B: Never assumes a and b share no common factor, so after correctly showing both are even the argument has nothing left to contradict; two numbers with even representations is not on its own impossible, and choosing a/b in lowest terms is exactly what is needed to make 'both even' an actual contradiction.
   • Why not C: Computes (2c)^3 as 2c^3 instead of 8c^3, an arithmetic slip that loses the correct relation b^3 = 4c^3; this error breaks the chain of reasoning and leads to the wrong conclusion that the method fails, when the correct expansion in fact completes the proof.
   • Why not D: Justifies 'a^3 even implies a is even' by appealing to 'an even integer always has an even cube' - but that statement is the CONVERSE of the fact given in the question (which says a cube being even implies the original integer is even); citing the converse of a fact is not the same as citing the fact itself, even though, as here, the conclusion drawn happens to coincide with what the given fact would also justify.

4. Question 4Answer: C

   1. To disprove a universal claim about every positive integer n, it suffices to exhibit one positive integer n for which the expression is not prime.
   2. Write n^2 - n + 11 = n(n-1) + 11. At n = 11, this is 11 x 10 + 11 = 110 + 11 = 121.
   3. 121 = 11^2, so 121 is composite, not prime.
   4. n = 11 is a positive integer for which the expression fails to be prime, so it is a valid counterexample, matching option C; spotting n(n-1)+11 = 11 x 10 + 11 = 11 x 11 is much faster than trial-dividing 121 from scratch, and faster still than testing every n up to 11 by brute force.
   • Why not A: Tests n=10 and, having found no small prime factor, wrongly concludes 101 'must be composite'; not finding a small factor is evidence of primality, not compositeness, and 101 is in fact prime, so this value does not disprove anything.
   • Why not B: Reaches the right value of n but miscomputes the arithmetic: 121 - 11 + 11 equals 121 (since -11+11=0), not 110; the claimed value of 110 is simply wrong, so this option does not correctly evaluate the expression it is trying to use as a counterexample.
   • Why not D: Uses n=0, which is not a positive integer and so falls outside the domain of the conjecture; and even setting that aside, 11 is prime (a number is composite only if it has a factor other than 1 and itself, and '11 x 1' does not show that), so this value would not disprove the claim even if it were admissible.

5. Question 5Answer: D

   1. From statement (2), 'if x is a multiple of 9, then the digit sum is a multiple of 9', and statement (3), 'the digit sum is not a multiple of 9', modus tollens gives: x is not a multiple of 9.
   2. Now apply modus tollens again, this time to statement (1), 'if x is a multiple of 18, then x is a multiple of 9', using the fact just derived that x is not a multiple of 9: this gives x is not a multiple of 18.
   3. Each step uses a different one of the two given conditional statements together with the negation of its conclusion, chaining two applications of modus tollens rather than stopping after one.
   4. So the fully justified deduction is that x is not a multiple of 18, matching option D.
   • Why not A: Correctly reaches (via statements (2) and (3)) that x is not a multiple of 9, but then wrongly stops there; statement (1) says multiples of 18 are always multiples of 9, so its contrapositive - x not a multiple of 9 means x is not a multiple of 18 - is available and gives a further, stronger conclusion that this option misses.
   • Why not B: Correctly derives that x is not a multiple of 9, but then wrongly claims this 'tells us nothing' about x and 18 and defaults to concluding x IS a multiple of 18; in fact statement (1) says multiples of 18 are always multiples of 9, so its contrapositive says non-multiples of 9 are never multiples of 18, which is the opposite of what this option concludes.
   • Why not C: Directly contradicts statement (3), which states the digit sum is NOT a multiple of 9; this option does not deduce anything new, it simply negates a given premise.

6. Question 6Answer: B

   1. Listing the sums 1, 1+8=9, 1+8+27=36, 1+8+27+64=100 suggests the pattern 1^2, 3^2, 6^2, 10^2, that is, the squares of the triangular numbers, leading to the conjecture S(n) = (n(n+1)/2)^2.
   2. To justify this for every n at once, compare consecutive sums: S(n) - S(n-1) should equal n^3, the cube being added at step n.
   3. Algebraically, S(n) - S(n-1) = (n(n+1)/2)^2 - ((n-1)n/2)^2 = (n^2/4)[(n+1)^2 - (n-1)^2] = (n^2/4)(4n) = n^3, exactly as required.
   4. Since S(0) = 0 correctly starts the pattern and each step adds precisely the next cube, S(n) = (n(n+1)/2)^2 is proved for every positive integer n, matching option B; this telescoping comparison is far quicker than trying to verify the formula case by case forever.
   • Why not A: States the correct conjecture, (n(n+1)/2)^2, but the justification only checks four small cases and then simply asserts the pattern 'must hold for every n' with no argument for why; agreement on finitely many cases suggests a conjecture but does not by itself justify it for all n.
   • Why not C: Proposes n(n+1)(2n+1)/6, which is actually the formula for the sum of the first n SQUARES, not cubes; substituting n=1,2,3,4 gives 1, 5, 14, 30, which does not match the observed sums 1, 9, 36, 100 at all, so the claimed agreement is false.
   • Why not D: Confuses the nth cube itself, n^3, with the SUM of the first n cubes; these are different quantities (the sum is a running total), and the sum is not equal to the last term added, just as it would be wrong to say the sum of the first n odd numbers equals the nth odd number itself.

7. Question 7Answer: A

   1. A direct proof of a 'for every integer n' statement should begin by introducing the variable and doing the algebra that follows from it: line W introduces n and expands (n+1)^3.
   2. The next step performs the subtraction that follows: line X computes (n+1)^3 - n^3 - 1 = 3n(n+1).
   3. Before that expression can be simplified further, line V establishes that n(n+1) is even and introduces m via n(n+1) = 2m, which line Y then substitutes to reach 6m.
   4. Finally line Z states the overall conclusion, giving the order W, X, V, Y, Z, which is option A; each line must only use quantities already introduced by an earlier line, and this is the only ordering among those given where that holds throughout.
   • Why not B: Opens with V, before W has introduced n as 'any integer' at all; V's claim about 'n and n+1' refers to a variable, n, that no earlier line has declared, so the very first line of the proof already refers to a quantity that has not yet been set up, exactly as in option D but with V instead of X.
   • Why not C: Places Y, which substitutes 'n(n+1) = 2m', before V, the line that actually defines m via n(n+1) = 2m; this uses m before it has been introduced, so the substitution in Y refers to a quantity that does not yet exist at that point in the proof.
   • Why not D: Opens with X, subtracting n^3 and 1 'from both sides of that equation', before W has stated the equation (n+1)^3 = n^3 + 3n^2 + 3n + 1 being subtracted from, and before n has even been introduced; the algebraic step cannot precede the expansion that defines the equation it manipulates.

8. Question 8Answer: C

   1. c is odd, so c^2 is odd (an odd number squared is always odd).
   2. Since a^2 + b^2 = c^2, a^2 + b^2 must also be odd. A sum of two squares is odd exactly when one square is odd and the other even, since odd+odd=even and even+even=even.
   3. A square is odd exactly when the original number is odd, and even exactly when the original number is even (squaring preserves parity), so exactly one of a, b is even and the other is odd.
   4. This holds for every such triple with c odd, not just specific examples, so the answer is C; checking every combination of parities directly, rather than guessing from one example triple, is what makes this a genuine proof rather than an observation.
   • Why not A: If both a and b were even, a^2+b^2 would be even + even = even, so c^2 would be even and hence c would be even (since a square is even only when the original number is even); this contradicts the given fact that c is odd, so both cannot be even.
   • Why not B: If both a and b were odd, a^2+b^2 would be odd + odd = even, so again c^2 would be even and c would be even, contradicting c odd; both cannot be odd either.
   • Why not D: Treats the parity of a and b as unresolved or dependent on the specific triple chosen, but the argument rules out 'both even' and 'both odd' outright, for every triple with c odd, not just some of them, leaving 'exactly one of each' as forced in every case.

9. Question 9Answer: B

   1. To disprove 'if n is prime then 2^n-1 is prime', a valid counterexample needs n itself to be prime, with 2^n-1 not prime.
   2. Take n = 11, which is prime. Then 2^11 = 2048, so 2^11 - 1 = 2047.
   3. 2047 = 23 x 89, both greater than 1, so 2047 is composite, not prime.
   4. n = 11 satisfies the hypothesis (prime) but not the conclusion (2^n-1 prime), so it is a valid counterexample, matching option B; this is the smallest prime n for which the pattern seen at n=2,3,5,7 first breaks down.
   • Why not A: Uses n=9, but 9 = 3 x 3 is not prime, so it falls outside the domain of the claim, which only concerns prime n; the claim can only be disproved by a value of n that is itself prime, as 11 is and 9 is not.
   • Why not C: Claims 8191 is even, but 2^13 is even, so 2^13 - 1 is odd (in fact 8191 is odd, and is itself prime, so it would not disprove the claim even if the parity claim were corrected); 2^n - 1 is always odd for every positive integer n, since it is one less than an even number.
   • Why not D: Miscalculates 2^11 - 1 as 2^(11-1) = 2^10 = 1024, treating 'subtract 1 from the exponent' as if it were the same as 'subtract 1 from the value'; the correct value is 2^11 - 1 = 2048 - 1 = 2047, not 1024.

10. Question 10Answer: D

    1. Factorising, x^2 - 5x + 6 = (x-2)(x-3), so P is true exactly when x = 2 or x = 3.
    2. If Q holds (x=2), then x=2 is one of the two roots, so P certainly holds; so Q is sufficient for P.
    3. But P can hold without Q: x=3 satisfies the equation without equalling 2, so P does not guarantee Q, meaning Q is not necessary for P.
    4. So Q is sufficient but not necessary for P, matching option D; spotting the second root x=3 from the factorisation is what settles necessity, without it one might wrongly assume the two statements are equivalent.
    • Why not A: Would require P and Q to be equivalent (the equation holds exactly when x=2), but x=3 also satisfies P without satisfying Q, so P does not imply Q; the two are not equivalent, and Q is not necessary for P.
    • Why not B: Reverses which direction holds: Q being necessary for P would require that P implies Q, that is, every solution of the equation equals 2; but x=3 is a solution of P that is not equal to 2, so P does not imply Q, meaning Q is not necessary for P.
    • Why not C: Misses that x=2 does satisfy the equation, so Q does imply P: Q being true (x=2) always makes P true, which is exactly what makes Q sufficient for P, so 'neither' understates the relationship that does hold.

11. Question 11Answer: C

    1. Assume, for contradiction, that log_2(3) = a/b for positive integers a and b, the negation of 'log_2(3) is irrational'.
    2. Then 2^(a/b) = 3, and raising both sides to the power b gives 2^a = 3^b.
    3. But 2^a is even for every positive integer a, while 3^b is odd for every positive integer b, since an odd number raised to any power remains odd; an even number can never equal an odd number.
    4. This contradiction shows the assumption was false, so log_2(3) cannot be written as a/b for positive integers a and b, and is irrational, matching option C; this parity clash is far quicker than any argument based on prime factorisation uniqueness.
    • Why not A: Assumes the very statement to be proved, that log_2(3) is irrational, rather than its negation; a contradiction proof must start from 'log_2(3) is rational' (nameable as a/b) and derive an impossibility, not restate the target as already established.
    • Why not B: Claims 3^b is even for every positive integer b, but 3 is odd and an odd number raised to any power stays odd, so 3^b is always odd, never even; this false premise is exactly backwards and removes the parity clash that actually makes the contradiction work.
    • Why not D: Tests only the single pair a=2, b=1 and, finding 2^2 not equal to 3^1, treats this as disproving the assumption; a contradiction proof needs to rule out EVERY possible pair of positive integers a, b, not just one arbitrarily chosen pair, since the assumption claims some pair works, not that this specific pair does.

12. Question 12Answer: A

    1. Complete the square: x^2 - 6x + 11 = (x-3)^2 - 9 + 11 = (x-3)^2 + 2.
    2. Since a square can never be negative, (x-3)^2 >= 0 for every real number x, so (x-3)^2 + 2 >= 0 + 2 = 2.
    3. So x^2 - 6x + 11 >= 2 for every real x, and since 2 > 0, x^2 - 6x + 11 > 0 for every real x.
    4. This single algebraic rearrangement settles the inequality for every real number at once, matching option A, unlike checking finitely many values or an incomplete discriminant argument that never pins down the sign.
    • Why not B: Claims x^2 - 6x + 11 = (x-3)^2 + 11, but expanding (x-3)^2 + 11 gives x^2 - 6x + 9 + 11 = x^2 - 6x + 20, not x^2 - 6x + 11; the correct constant to add after completing the square is 11 - 9 = 2, not 11 itself, so this drops the necessary adjustment for the 9 introduced by (x-3)^2.
    • Why not C: Correctly shows the quadratic has no real roots, but 'never zero' by itself does not distinguish 'always positive' from 'always negative'; the argument also needs one value where the sign is known (for example, at x=0 the expression equals 11, which is positive) or the fact that the leading coefficient is positive, neither of which is mentioned here.
    • Why not D: Checks only three specific real numbers and concludes the inequality holds for 'every real number x'; unlike a finite check over integer cases, there are infinitely many real numbers between any two checked values, so three examples give essentially no information about the rest.