Grade 12 math puzzles — free practice & printable teacher quizzes

• What does the matching real part tell you about the difference of the two squares involving 1?
• Use that difference and |z| to find z² before expanding the expression you need.

Answer: A — −10

The difference (z + 1)² − (z − 1)² equals 4z. Since the two squares have the same real part, z has real part 0. With |z| = 2, this gives z = ±2i and z² = −4. Thus (z + i)² + (z − i)² = 2z² + 2i² = −8 − 2 = −10. The tempting −6 results from using i² = 1 in the last step.

• Follow what the equation forces as you move from one number to the next around the circle.
• If a = f(0), express f(x) in terms of a, then ask when f(x) could equal x.

Answer: D — 1013

Write a = f(0). Since applying g to an input moves it forward while applying g⁻¹ to an output moves it backward, the equation forces f(x) ≡ a − x (mod 2026). Every value of a gives a permutation. A fixed point would satisfy 2x ≡ a (mod 2026), which is possible exactly when a is even. There are 1013 odd values of a, so the answer is 1013. The tempting answer 2026 counts the permutations that do have fixed points as well.

• Look for a relationship among the three coordinate differences before expanding the dot product.
• Let x=a−b, y=b−c, and z=c−a. Use (x+y+z)² to relate their pairwise products to their squares, then use the given information to find the sum of those squares.

Answer: D — −13

Set x=a−b, y=b−c, and z=c−a. Since x+y+z=0, the dot product is xy+yz+zx=−(x²+y²+z²)/2. Also, x²+y²+z²=3(a²+b²+c²)−(a+b+c)²=42−16=26. Thus u·v=−13. The tempting −26 comes from finding the correct sum of squares but forgetting to divide by 2.

• As a changes, where could two solutions merge? The quadratic’s zeros divide the graph into regions worth considering separately.
• For x > 0, compare |x² − 4x + 3|/x with a, and check that any candidate lies in the region where you found it.

Answer: D — 4 − 2√3

Since a > 0, any solution has x > 0. On (0, 1) and (3, ∞), the equation is x² − (4 + a)x + 3 = 0. This quadratic changes sign between 0 and 1 and again between 3 and ∞, so it has exactly one root in each interval. For 1 < x < 3, the equation becomes a = 4 − x − 3/x. This expression rises from 0 to its maximum 4 − 2√3 at x = √3, then falls to 0; it equals a at exactly one point in this interval only when a = 4 − 2√3. That gives three solutions total. The tempting 4 + 2√3 also makes the middle quadratic’s discriminant zero, but its double root is −√3, outside the middle interval.

• Try to make the expression symmetric in x and 1/x before solving for x.
• Combine the logarithms, then factor the cubic expressions and look for x² + 1/x².

Answer: A — log 8

Combining the logarithms gives log(((x³ − 1)(1 + x⁻³))/((x − 1)(1 + x⁻¹))). Factoring and cancelling reduces the argument to ((x² + x + 1)(x² − x + 1))/x² = x² + 1 + x⁻². This equals (x + 1/x)² − 1 = 8, so the answer is log 8. The tempting log 9 comes from overlooking the −1 in that final identity.

• Look at the numbers of white beads between successive black beads. What happens to that list when you choose a different black bead as your starting point?
• Count the positive four-part sums, then check which lists remain unchanged when shifted by one, two, or three places.

Answer: B — 10

The four gaps between successive black beads contain positive numbers of white beads summing to 8, giving C(7,3) = 35 ordered lists. Under cyclic shifts of a list, 35 lists are unchanged by a shift of zero places; 1 is unchanged by a shift of one place; 3, of the form (a,b,a,b) with a+b=4, are unchanged by a shift of two places; and 1 is unchanged by a shift of three places. Each distinct necklace contributes four across these fixed-list counts, so the answer is (35+1+3+1)/4 = 10. The tempting answer 35 counts the same necklace again when a different black bead is chosen as the starting point.

• Express the two block sums using the total of the first several odd integers. Then ask whether a solution could lead to a smaller one.
• Once you have N² − 2k² = 1, try replacing (N, k) with (3N − 4k, 3k − 2N).

Answer: A — 577

If the first block contains k numbers, its sum is k² and the total is N². Thus N² − 2k² = 1. Every solution with N > 3 yields a smaller nonnegative solution (3N − 4k, 3k − 2N) to the same equation; reversing this step gives the successive solutions (N, k) = (1, 0), (3, 2), (17, 12), (99, 70), (577, 408). So 577 is the first possible N above 100. The tempting 239 reverses the required difference: 239² − 2·169² = −1.

• Think about how the cubic factors at each root: its other two roots give the denominator a useful meaning. The root equation can also tame the eighth powers.
• Reduce x⁸ modulo the cubic to a quadratic. Then write that quadratic using its values at a, b, and c, and compare coefficients of x².

Answer: B — 28

Write f(x) = (x − a)(x − b)(x − c). At any root r, f′(r) = 3r² − 3 is the product of r’s differences from the other two roots. Using r³ = 3r − 1 repeatedly gives r⁸ = 28r² − 27r + 6. Let R(x) = 28x² − 27x + 6. Interpolating R from its values at a, b, and c, the coefficient of x² is R(a)/f′(a) + R(b)/f′(b) + R(c)/f′(c). The requested sum is therefore 28. The tempting 186 is the sum a⁸ + b⁸ + c⁸; it ignores the denominators.

• Try checking fourth-power remainders using the prime factors of 65 rather than calculating large powers.
• For a base not divisible by 13, Fermat’s little theorem tells you what its twelfth power is modulo 13.

Answer: B — 56

The remainders 16, 26, 35, and 61 occur for bases 2, 13, 20, and 4, respectively. But if x⁴ ≡ 56 (mod 65), then x⁴ ≡ 4 (mod 13), so 13 does not divide x. Fermat’s little theorem would give 1 ≡ x¹² ≡ 4³ ≡ −1 (mod 13), a contradiction. The trap is that 4⁴ = 256 ends in 56, but its remainder modulo 65 is 61.

• The repeated bar is not equally likely to be any of the four bars: its size also affects which choices remain for the other two students.
• For each possible repeated bar, count pairs from that bar, then choose the other two students from two different bars.

Answer: A — 303/724

Count each four-student group once, according to its repeated bar. The numbers of valid groups when that bar is under 2 hours, 2 to under 4 hours, 4 to under 6 hours, and 6 hours or more are, respectively, 136,500; 337,125; 270,000; and 70,875. Thus the conditional probability is (270,000 + 70,875)/(136,500 + 337,125 + 270,000 + 70,875) = 303/724. The tempting 1/2 treats all four bars as equally likely to supply the repeated pair, but groups with different repeated bars occur in different numbers.

• Draw a few rows, especially rows numbered by powers of two. Look for a way to skip ahead instead of building every row.
• Represent the lit LEDs in a row by the nonzero coefficients of a polynomial modulo 2. Moving down one row multiplies that polynomial by 1+x.

Answer: C — 256

The polynomial for row 2026 is (1+x)^2026, with coefficients taken modulo 2. For any power of two, (1+x)^(2^k) = 1+x^(2^k) modulo 2. Since 2026 = 1024+512+256+128+64+32+8+2, the polynomial is a product of eight such factors. Choosing one term from each factor gives 2^8 = 256 distinct exponents, so 256 LEDs are lit. The tempting 2027 counts every position in the row as lit, overlooking LEDs toggled an even number of times.

• Reflecting a path across an axis can turn a bend at that axis into part of a straight-line comparison. How could this help with two required visits?
• Reflect the start across the y-axis and the finish across the x-axis. Check whether the segment joining the reflected points crosses the axes in the required order.

Answer: D — 10√2

Let P be the visit to the y-axis and Q the later visit to the x-axis. Reflecting the start to (−2,7) and the finish to (8,−3) preserves the lengths of the first and last parts of the route. The route is therefore at least as long as the straight-line distance between those reflected points: √(10² + 10²) = 10√2. The joining segment crosses the y-axis at (0,5) and then the x-axis at (5,0), so a route of length 10√2 is possible within the closed first quadrant. The value 2√29 comes from reflecting only the start across the y-axis and measuring to (8,3); that straight segment never reaches the x-axis, so it does not satisfy the second required visit.