Daily Math Puzzle: 2026-10-14
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2026-10-14
At Professor Alistair's grand magic show, he presents a magical top hat. Initially, 25% of the coins inside the hat are pure gold. For his grand finale, Professor Alistair dramatically removes 3 gold coins from the hat. Then, with a flourish, he adds 9 silver coins. After these magical manipulations, only 10% of the coins in the hat are now pure gold. Assuming the number of coins must always be a whole number, what was the minimum total number of coins in the hat *before* Professor Alistair began his final trick?
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Solution
24 — Let 'C' be the initial total number of coins in the hat.
Initially, 25% of the coins are gold, so the number of gold coins is 0.25C.
Step 1: Professor Alistair removes 3 gold coins.
Number of gold coins now: 0.25C - 3.
Total number of coins now: C - 3.
Step 2: He adds 9 silver coins.
Number of gold coins remains: 0.25C - 3 (since only silver coins were added).
Total number of coins now: (C - 3) + 9 = C + 6.
After these manipulations, 10% of the coins are gold. So, we can set up the equation:
(Number of gold coins) / (Total number of coins) = 10%
(0.25C - 3) / (C + 6) = 0.10
Now, we solve for C:
0.25C - 3 = 0.10 * (C + 6)
0.25C - 3 = 0.10C + 0.6
0.25C - 0.10C = 0.6 + 3
0.15C = 3.6
C = 3.6 / 0.15
C = 360 / 15
C = 24
We must also ensure that the initial number of gold coins (0.25C) is a whole number, as you cannot remove a fraction of a coin. If C = 24, then initial gold coins = 0.25 * 24 = 6. This is a whole number, and since 24 is the smallest integer value that satisfies the equation and makes 0.25C a whole number (C must be a multiple of 4), it is the minimum total number of coins.
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