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805+ Level|   Math Related|         
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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 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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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 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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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 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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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 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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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 11 Jul 2024, 07:00
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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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for the GMAT Olympics 2024

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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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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 13 Aug 2025, 13:27
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