Math Problems and Logic Puzzles

This is the two-box paradox. The tempting reasoning that switching gives $12.50 expected value is flawed. When you see $10, the possible pairs are ($5,$10) or ($10,$20). Since you chose the $10 box, if the pair was ($5,$10) you definitely picked the larger one (probability 1), but if the pair was ($10,$20) you only had a 50% chance of picking the smaller one. This creates selection bias: seeing $10 is more likely from the ($5,$10) pair (2/3) than from ($10,$20) (1/3). So the other box contains $5 with 2/3 probability and $20 with 1/3 probability, giving expected value of $10 - no gain from switching.

Let N be the number of inhabitants. Each has a unique number of hairs. If the max hairs is H, then N > H. If no one has 40360 hairs, and each has a unique count, the counts could be 0, 1, ..., 40359, 40361... To satisfy N > H, the max possible N is 40360 (with hair counts 0 to 40359).

Statement E (none of the girls occupy adjacent cottages) can be derived from statements B and D. Statement B says Eliza and Anne won't stay beside occupied cottages, and statement D places one vacant house between Anne and Jack. Combining these with the gender constraints ensures the three women (Eliza, Anne, Karen) cannot be adjacent to each other.

Based on the cottage arrangement: [Vacant, Tom, Eliza, Vacant, Anne, Vacant, Jack, Karen, Peter], the occupied cottages next to a vacant cottage are Tom, Eliza, Anne, and Jack. This gives a total of 4 people, making 4 the correct answer.

Both N (natural numbers) and Z (integers) are countably infinite sets with the same cardinality (aleph-null). A bijection exists between them: 0↔0, 1↔1, 2↔-1, 3↔2, 4↔-2, 5↔3, 6↔-3, etc., proving they have the same number of elements despite Z appearing 'larger'.

By systematically applying all six constraints (no ends for Anne/Tom/Jack, Eliza/Anne beside occupied cottages, Karen next to Peter and Jack, one vacant between Anne and Jack, girls not adjacent, Tom beside end cottage), we can determine the valid cottage arrangements. Statement 1 (Anne between Eliza and Jack) and Statement 3 (Tom beside Peter) are true, while Statement 2 about four persons having occupied cottages on either side is incorrect.

This sequence follows a digit-based pattern where for each three-digit number ABC, specific relationships between digits determine the sequence progression. The pattern alternates through different digit operations to generate 745 and 999 as the next two terms.

Using Bayes' theorem or simple counting: there are 3 white marbles total, and 2 of them are in Bag A (where the remaining is white). If you drew a white marble, the probability it came from Bag A is 2/3. Only Bag A guarantees a white remaining; Bag C has a black remaining. Therefore the probability is 2/3.

Factor out 3 from the second expression: 6x - 3y = 3(2x - y). Since we're given that 2x - y = 4, substituting gives 3 × 4 = 12.

We are given $x = y = 2z$, which means $y = x$ and $z = x/2$. Substituting these into $xyz = 256$ gives $x \times x \times (x/2) = 256$, which simplifies to $x^3 = 512$. Taking the cube root of both sides yields $x = 8$. Distractors represent incorrect algebraic substitutions.

(1/10)^18 - (1/10)^20 = 10^(-18) - 10^(-20) = 10^(-20)(10^2 - 1) = 10^(-20) × 99 = 99/10^20. Option A represents this as 99/10^20 (where 1020 denotes 10 to the power of 20).

Combined rate = 1/20 + 1/30 - 1/40 = (6 + 4 - 3) / 120 = 7/120. The time required is the reciprocal of the rate: 120 / 7 = 17 1/7 minutes.

Let the 5-digit number be N. If we put 1 at the beginning, we get 100000 + N. If we put 1 at the end, we get 10N + 1. According to the problem: (10N + 1) = 3 × (100000 + N). Solving: 10N + 1 = 300000 + 3N → 7N = 299999 → N = 42857. Verification: 142857 ÷ 428571 = 1/3. The other options don't satisfy this mathematical relationship.

The sequence is generated by the formula $n^2 + n$ for $n = 1, 2, 3, 4, 5$. Specifically: $1^2+1=2$, $2^2+2=6$, $3^2+3=12$, $4^2+4=20$. The next term for $n=5$ is $5^2+5=30$. Distractors represent incorrect pattern assumptions.

To maximize the product of two numbers formed by digits 1-9, we distribute the largest digits to the highest place values alternately: 9 and 8, then 7 and 6, and so on. Comparing the products, $9642 \times 87531 = 843,973,902$, which is the maximum possible product.

If Leonardo has 0 Rembrandts, then Pablo is wrong (0 is not at least 4), Henri is wrong (0 is not at least 1), and Edvard is right (0 is less than 4). Since only one friend is right, this scenario is perfectly consistent. Any other number of paintings would make at least two friends correct.

If Richard tells the truth on Tuesday, Day 1 (Monday) is a lie, so his statement 'I lie on Monday and Tuesday' is false (since he tells the truth on Tuesday). Day 2 (Tuesday) is the truth, and his statement 'Today is Thursday, Saturday, or Sunday' is a lie. Day 3 (Wednesday) is a lie, so his statement is false. This perfectly fits the rules.

In this truel, C has the lowest accuracy (50%) but the optimal strategy: firing into the air on his first turn. This forces the deadlier A (100%) and B (80%) to target each other first. Once one is eliminated, C gets the first shot at the survivor, giving him a survival probability of over 50%.