Brainteasers interview questions

The key insight is that when two ants meet and reverse direction, the system is equivalent to the ants passing through each other without interaction, because the ants are indistinguishable. Under this "pass-through" model, each ant simply continues in its original direction forever. Therefore, the number of ants that reach Priya's end equals the number of ants that were initially moving leftward (toward Priya), and the number that reach Theo's end equals the number initially moving rightward (toward Theo).

Initially, all 40 ants at Priya's end move rightward (toward Theo), and all 80 ants at Theo's end move leftward (toward Priya). Under the pass-through equivalence, the ants that reach Priya are those that started at Theo's end and moved leftward: 80 ants. The ants that reach Theo are those that started at Priya's end and moved rightward: 40 ants. Thus $x = 80$, $y = 40$, and

More formally, let $x$ be the number reaching Priya and $y$ the number reaching Theo. Under the pass-through model, each ant's trajectory is a straight line from its start to the opposite end. Since all ants move at the same speed, the number reaching each end is exactly the number that started at the opposite end. Hence $x = 80$, $y = 40$, and the difference is $40$.

Let the statements be numbered $1,2,\dots,100$. Statement $k$ says: "at most $k-1$ of these $100$ statements are true."

Let $T$ be the total number of true statements. For statement $k$ to be true, we must have $T \le k-1$; for it to be false, $T > k-1$. Hence statement $k$ is true exactly when $k \ge T+1$.

Thus the true statements are those with indices $T+1, T+2, \dots, 100$. The number of such statements is $100 - T$. But this number must equal $T$ itself, because $T$ is the total number of true statements. Therefore:

Check: If $T=50$, then statements $51$ through $100$ are true (50 statements). Statement $51$ says "at most $50$ are true" — true because exactly $50$ are true. Statement $50$ says "at most $49$ are true" — false because $50 > 49$. All statements $1$ through $49$ are false because each claims "at most $k-1$ are true" with $k-1 \le 48$, but $50 > k-1$. All statements $52$ through $100$ are true because each claims "at most $k-1$ are true" with $k-1 \ge 51$, and $50 \le k-1$. Hence exactly $50$ statements are true, consistent with $T=50$.

Thus the answer is $50$.

We determine the outcome by backward induction on the number of tigers, $n$. The key observation is that a tiger will eat the sheep only if doing so guarantees its own survival after it turns into a sheep.

• $n=1$: The lone tiger eats the sheep, becomes a sheep, and faces no remaining tigers. It survives, so the sheep is eaten.
• $n=2$: If a tiger eats the sheep, it becomes a sheep, leaving one tiger and one sheep. By the $n=1$ case, that remaining tiger will eat the new sheep, so the first tiger would die. Since tigers value survival, neither will eat. The sheep survives.
• $n=3$: If a tiger eats, it becomes a sheep, leaving two tigers and one sheep. From $n=2$, the two tigers will not eat the sheep (they would die), so the eating tiger survives. Hence some tiger will eat, and the sheep is eaten.
• $n=4$: Eating leads to the $n=3$ case, where the sheep is eaten, so the eating tiger would die. No tiger eats; the sheep survives.

Continuing this logic, the sheep is eaten if and only if $n$ is odd. For $n=100$ (even), no tiger will eat the sheep.

This is a classic "jeep problem" (desert crossing) variant. The tanker starts with $3000$ gallons at Port A, must travel $1000$ miles to Port B, loses $1$ gallon per mile, and has capacity $1000$ gallons. The goal is to maximize delivered oil.

Key insight: The optimal strategy uses intermediate depots and shuttles oil forward in stages, each stage moving a certain amount of oil a certain distance while consuming fuel for round trips. The number of trips decreases by one each stage.

Let $D = 1000$ miles, capacity $C = 1000$ gallons, initial fuel $F = 3000$ gallons. Consumption is $1$ gallon per mile.

General principle: To move $n$ full loads (i.e., $n \times 1000$ gallons) a distance $d$, you make $n$ forward trips and $n-1$ return trips, consuming $(2n-1)d$ gallons. The amount delivered forward is $nC - (2n-1)d$.

Stage 1: Start with $3000$ gallons at A, i.e., $n_1 = 3$ loads. Move forward distance $d_1$ such that after shuttling we have exactly $n_2 = 2$ loads ($2000$ gallons) at the next depot. Fuel consumed: $(2\cdot3-1)d_1 = 5d_1$. Remaining: $3000 - 5d_1 = 2000 \Rightarrow d_1 = 200$ miles.

Stage 2: At mile $200$, we have $2000$ gallons ($n_2 = 2$ loads). Move forward distance $d_2$ to reduce to $1$ load ($1000$ gallons). Fuel consumed: $(2\cdot2-1)d_2 = 3d_2$. Remaining: $2000 - 3d_2 = 1000 \Rightarrow d_2 = 333\frac{1}{3}$ miles.

Stage 3: At mile $200 + 333\frac{1}{3} = 533\frac{1}{3}$, we have $1000$ gallons ($n_3 = 1$ load). Drive directly to B, which is $1000 - 533\frac{1}{3} = 466\frac{2}{3}$ miles away. Fuel consumed: $466\frac{2}{3}$ gallons. Delivered: $1000 - 466\frac{2}{3} = 533\frac{1}{3}$ gallons.

Thus the maximum deliverable is $533\frac{1}{3}$ gallons, which rounds to $533$ gallons.

Verification: Total fuel consumed = $5\times200 + 3\times333\frac{1}{3} + 1\times466\frac{2}{3} = 1000 + 1000 + 466\frac{2}{3} = 2466\frac{2}{3}$ gallons, matching $3000 - 533\frac{1}{3}$. Total distance traveled = same sum, confirming consumption rate.