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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially

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Cocasocola

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11 Jul 2024, 06:06

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x = (1/6)^p
y = (1/6)^q

1/x + 1/y = 7992
we have
6^p + 6^q = 7992 = 6^3 * 37 = 6^3 + 6^5
Because x < y, then P > Q

Answer: P = 5, Q = 3

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11 Jul 2024, 06:25

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Bunuel

The king of Trishkinda has specially designed an ivory dice. Six exotic stones – agate, jasper, chalcedony, sapphire, topaz, garnet – are embedded on the six sides of the dice, such that one side has exactly one type of stone and no two sides have the same type of stone. When the dice is rolled, the probability of the agate stone appearing is $\frac{1}{6}$. Let there be values P, Q, x, y such that when the dice is thrown P times, the probability of the agate stone appearing all P times is x and when the dice is thrown Q times, the probability of the agate stone appearing all Q times is y. Furthermore, $x < y$ and $\frac{1}{x} + \frac{1}{y} = 7,992$.

In the table, select a value for P and a value for Q that are jointly consistent with the given information. Make only two selections, one in each column.

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11 Jul 2024, 06:33

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X = (1/6)^P
Y = (1/6)^Q

(1/X) + (1/Y) = 7992 As given

From this, we can calculate 6^P + 6^Q = 7992

As X<Y, so P>Q
Testing for values of P & Q, 7776+216 = 7992
Therefore, P=5, Q=3

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11 Jul 2024, 06:38

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x = (1/6)^p
y = (1/6)^q

y>x , which means that p > q

1/x + 1/y = 7992
substituting x and y

6^p + 6^q = 7992

as we know that p > q , and sum is greater than 7000 by hit and trial we get p = 5 and q= 3

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11 Jul 2024, 07:00

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Bunuel

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we know,
x < y
thus, $P > Q$

$x = \frac{1}{6}^P$
$y = \frac{1}{6}^Q$

$6^P + 6^Q = 7992$
$6^Q * (6^P-Q + 1) = 7992$

Solving above equation, we get
$Q = 3$
$P-Q = 2$ => $P = 5$

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13 Aug 2025, 13:27

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