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Home » Broo Hacks Every Morning—People Are Losing Their Minds Over These Simple Tricks! - Redraw

Broo Hacks Every Morning—People Are Losing Their Minds Over These Simple Tricks! - Redraw

📅 May 6, 2026 👤 scraface

May 06, 2026

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📌 Solution: The code fixes 4 digits (D, E, F) and permutes only A, B, C in the remaining 3 positions. The total number of permutations of A, B, C is $3! = 6$. Among these, the number of permutations where A comes before B is exactly half, since in half the arrangements A precedes B and in half B precedes A (the cases are symmetric).

📌 Thus, number of valid codes where A comes before B is:

📌 Since the outer digits are fixed, this counts the number of valid codes satisfying the constraint.

📌 Combine the results from steps 1 and 2:

📌 Next, find $ xy $. Use the identity:

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Bottle 1: capacity 729, we put 729 ml → full.
Remaining: 0 → done.
But then only one bottle.
Unless the "largest" is not 729, but the sequence starts with capacity...
Alternative: The first bottle used has capacity 729, but we only fill...
Given ambiguity, assume standard geometric series: total volume to be...
But total is \(S = a / (1 - r) = a / (1 - 1/2) = 2a\)
Set \(2a = 729\) → \(a = 364.5\)
Then bottles: 364.5, 182.25, 91.125, 45.5625, 22.78125, 11.390625, 5.6...
Sum after n terms: \(S_n = a(1 - (1/2)^n)/(1 - 1/2) = 2a(1 - 2^{-n})\)

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