Math & Logic Puzzles

This is a survival logic puzzle. Lions cannot survive 3 months without food - they would starve to death. Room 3 is actually the safest because the lions are dead. Rooms 1 and 2 pose obvious lethal threats (fire, assassins). Room 4 is explicitly dangerous with killers searching. The riddle tests whether you recognize biological impossibility.

This question tests spatial reasoning in races. If you overtake the last runner, you must have been lapped or there's a logical contradiction - you can't be behind the last person if you're passing them. Therefore the statement is false. The phrasing '3 were running' suggests a race context where overtaking the last runner is impossible.

For 101^100 - 1, we can write it as (100+1)^100 - 1. Using binomial expansion, all terms except the constant term are multiples of 100, so the expression is divisible by 100. For higher powers of 10, we need to check: 10000 = 10^4 divides it because the binomial expansion includes terms with 10^4 as a factor.

91+92+93+94+95+96+97+98 = 703, and 703/6 = 117 remainder 1, so remainder is 1, not 0. Alternatively, note that consecutive integers have remainders cycling 1,2,3,4,5,0. Summing eight consecutive numbers gives remainders summing to 0 (if sum of 6 consecutive numbers). Checking: 91+98=189 (divisible by 3), 92+97=189, 93+96=189, 94+95=189. Total 756, which is divisible by 6, so remainder should be 0. But wait, 91+92+93+94+95+96+97+98 = 703? No, that's wrong. Let me recalculate: (91+98)+(92+97)+(93+96)+(94+95) = 189+189+189+189 = 756. 756/6 = 126 exactly. So remainder is 0. The claimed answer is correct.

For the quadratic equation 2x^2 + 2(p+1)x + p = 0, the discriminant is D = 4(p+1)^2 - 8p = 4(p^2+2p+1-2p) = 4(p^2+1). Since p^2+1 is always positive for all real p, D > 0 always. Therefore, the roots are always real and distinct.

Using the right triangles ABD and ACD, we find BC = BD + DC. From Pythagoras: BD = sqrt(17.5^2 - 3^2) and DC = sqrt(9^2 - 3^2). This gives BC ≈ 25.725. The circumradius formula R = abc/(4 × Area) with Area = (1/2) × BC × AD yields R = (17.5 × 9 × 25.725)/(4 × 38.588) ≈ 26.25.

Parallel lines have the same coefficients for x and y. The given line 3x + 7y = 10 has slope -3/7. Substituting point (4, 8) into 3x + 7y = c gives c = 3(4) + 7(8) = 12 + 56 = 68. The parallel line through (4, 8) is 3x + 7y = 68.

Multiples of 7 from 200 to 400 are: 203, 210, ..., 399. This is an arithmetic series with first term 203, last term 399, common difference 7. Number of terms = (399 - 203)/7 + 1 = 28 + 1 = 29. Sum = (29/2) × (203 + 399) = 29 × 301 = 8729.

Pipe A fills in a hours, so rate = 1/a tank per hour. With the leak, effective rate = 1/(3a). Let leak rate = -1/x (negative because it empties). Then 1/a - 1/x = 1/(3a). Solving: 1/x = 1/a - 1/(3a) = 2/(3a), so x = (3/2)a hours.

If A gives B 20 points, when A scores the game total, B scores (total - 20). This means A:B speed ratio = total:(total - 20). Similarly, A:C = total:(total - 32) and B:C = total:(total - 15). Solving these ratios: total - 20 = total - 32 + total - 15 - total, which gives total = 100.

Sn has terms 3, 8, 15, 24, 35... where the k-th term is k²+2k. S20 = sum from k=1 to 20 of (k²+2k) = 2870 + 420 = 3290. S'n has terms 6, 11, 18, 27... where the k-th term is k²+2k+3. S'20 = sum from k=1 to 20 of (k²+2k+3) = 2870 + 420 + 60 = 3350. Therefore S20 - S'20 = 3290 - 3350 = -60.

Let cost price be C. After 40% discount on marked price, selling price = C. So 0.60 × Marked Price = C, which means Marked Price = C/0.60 = 1.6667C. The markup is (Marked Price - Cost Price)/Cost Price × 100 = (1.6667C - C)/C × 100 = 66.67%.

Trailing zeros in n! come from factors of 10 = 2×5. Since there are always more factors of 2 than 5, we count factors of 5. For n=125: floor(125/5)=25, floor(125/25)=5, floor(125/125)=1. Total = 25+5+1=31 zeros. For n=120: floor(120/5)=24, floor(120/25)=4 → 28 zeros. So 125 is the smallest.

To find 2^1001 mod 1001, note that 1001 = 7 × 11 × 13. Using the Chinese Remainder Theorem: 2^1001 ≡ 4 (mod 7), 2^1001 ≡ 2 (mod 11), and 2^1001 ≡ 6 (mod 13). The unique number satisfying all three congruences is 123. Option A is correct.

A number that is both a perfect square and perfect cube must be a perfect 6th power (LCM of 2 and 3 is 6). The 6th powers less than 1000 are: 1^6 = 1, 2^6 = 64, 3^6 = 729. Sum = 1 + 64 + 729 = 794. Option D is correct.