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

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JDEWERPEF

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10 Jul 2024, 13:12

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

1/((1/6)^p )+1/((1/6)^q) = 7,992

6^p + 6^q = 7,992

6^q * (6^p-q + 1) = 6^3 * (6^2 +1)

p=5 , q = 3

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Bunuel

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$x = (\frac{1}{6})^P$

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

$y = (\frac{1}{6})^Q$

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

We also know that $x<y$ and as we're dealing with fractions, that means that $P>Q$ (a higher value as the exponent will result in a lower number since the base is the same).

Furthermore, we are told that:

$\frac{1}{x}+\frac{1}{y} = 7992$

Subbing in our values, we get:

$\frac{1}{1/6^P} + \frac{1}{1/6^Q} = 7992 $

$6^P+ 6^Q= 7992 $

Now, let's look at the different values of $6^k$ under 7992.

$6^1=6$, $6^2=36$, $6^3=216$, $6^4=1296$, $6^5=7776$

$6^5 + 6^3 = 7992$ satisfies the equation so P is 5 and Q is 3.

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Let's make the unique prime number decomposition of 7992 = 2^3*3^3*37=6^5+6^3

the probability of the agate stone appearing all P times is x then , x=(1/6)^P --> 1/x = 6^P

the probability of the agate stone appearing all Q times is y then , y=(1/6)^Q--> 1/y = 6^Q

then, according to 1/x+1/y=7,992 then 6^P + 6^Q = 6^5+6^3
Since x<y then P>Q
we then can conclude that P=5 and Q=3

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x and y are the probabilities
means 1/x+1/y are their recipricals
since probability is at 1/6 for each throw, we need to get throws when added together gets the sum of 7792

6^x+6^y (x<y) = 7992
x = 3, y=5

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P=5
Q=3

P>Q
since x < y

so i/x+1/y=7992
ie, 6^p+6^q =7992

using options
we see
6^3(37)=7992
works so
P=5
Q=3

this method took me over 4.9 mins
too lengthy to work through options,
looking to learn a faster method of solving this

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Bunuel

This question was provided by Experts' Global
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X= (1/6)^P
Y= (1/6)^Q
If X<Y
Then P>Q

1/X+1/Y = 7992
Then
6^P+6^Q = 6^5+6^3

Then
P=5
Q=3

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Bunuel

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The probability of a stone appearing once, as given in the case of agate = 1/6
The probability of this happening P times in a row = x = (1/6)^P
The probability of this happening Q times in a row = y = (1/6)^Q

We have x<y, thus (1/6)^P < (1/6)^Q, since 1/6 < 1, it means P>Q.

Also, (1/x + 1/y)= 7992 or 6^P + 6^Q = 7992
We see that 6^5 = 7776 and 6^3 = 216, and 7776+216= 7992

The key here is to recognise which is larger P or Q, here P>Q, so P=5 and Q=3.

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We have :
$x=(1/6)^P$
$y=(1/6)^Q$

Which implies that

$6^P + 6^Q =7992$

By triying the variables in the table we find that the combination is 5 and 3
Since x<y => P>Q

P=5
Q=3

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The probability to get the agate stone is 1/6 and the probability to get the agate stone n times is (1/6)ˆn.
Therefore:

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

and x<y => P>Q

and $\frac{1}{x}+\frac{1}{y} = 7992 => 6^P + 6^Q = 6^5 + 6^3 $

Therefore P=5 and Q=3

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As x < y -> P > Q

x = 1 / 6^P
y = 1 / 6^Q

1/x + 1/y = 7992
6^P + 6^Q = 7992
6^Q (6^(P-Q) + 1) = 7992 = 2^3 * 3^3 * 37 = 6^3 * 37

Q = 3

6^(P-Q) + 1 = 37
6^(P-Q) = 36 = 6^2
P-Q = 2
P = Q + 2 = 5

P = 5
Q = 3

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The probability of the agate stone appearing is $\frac{1}{6}$

When the dice is thrown $P$ times, the probability of the agate stone appearing all $P$ times is $x$
$x = \frac{1}{6^P}$

When the dice is thrown Q times, the probability of the agate stone appearing all Q times is y:
$y = \frac{1}{6^Q}$

since $x$, it must be $P>Q$

Also given, $\frac{1}{x} + \frac{1}{y} = 7992$
$ = 6^P + 6^Q = 7992$

As per the option choices, $P$ and $Q$ can take values - 1,2,3,4,5,6

$6^1 = 6$
$6^2 = 36$
$6^3 = 216$
$6^4 = 1,296$
$6^5 = 7,776$
$6^6 = 46,656$

We can observe that $6^5 + 6^3$ gives 7992

since, it must be $P>Q$,

Correct Option: $P = 5, Q = 3$

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7992 = 6^5 (7776) + 6^3 (216)
so, x is 1/(6^5), y is 1/(6^3) as x<y

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P(agate) = 1/6.
P(agate p times) = 1/(6^p). = x
P(agate q times) = 1/(6^q). = y

Since x < y -> 1/(6^p) < 1/(6^q) i.e. 6^p > 6^q i.e. p > q
ALso 1/x + 1/y = 6^p + 6^q = 7992 i.e 6 ^ (p+q) = 7992 -> p+q < 5

Answer: P: 3 Q: 1

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We needed to find the two numbers that jointly gave us the correct answer.
For this, we could use the answer choices to solve it.
If we assume P=3, X would be 1/6*1/6*1/6 = 1/216. We substitute this in the equatin 1/x + 1/y = 7.992, so we get 216 + 1/y = 7.992.
We solve for y --> 1/y = 7.776 (we prime factorize this to realize its 6^5), so y = 1/6^5, and Q = 5.
The answer is P=3 and Q=5

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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 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 1/x+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.

probability of the agate stone appearing is 1/6
P times is x
Q times is y
x < y
so P would be more number of times because probability of 2 times (1/36) is agate will be greater than 3 times (1/216) [this will be useful later]

1/x+1/y=7,992
can be simpleifiled to: (6)^a + (6)^b = 7992
(6)^a + (6)^b = 37*6^3
(6)^a + (6)^b = 6^3 ( 6 ^2 + 1)
(6)^a + (6)^b = 6^5 + 6^3
a = 3 & b = 5

So dice was rolled 5 and 3 times. As mentioned ealier the P would be 5 and Q would be 3.

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10 Jul 2024, 21:19

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Quote:

When the dice is rolled P times, the probability of the agate stone appearing all P times is (1/6)^P, x=(1/6)^P

When the dice is rolled P times, the probability of the agate stone appearing all Q times is (1/6)^Q, y=(1/6)^Q

Given x<y and 1/x+1/y=7992

So that means 6^P+6^Q=7992

6^P+6^Q=6^3*37

so P=5 and Q=3

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Given the probability of getting the right side of the dice is 1/6, we can define:
$x = \frac{1}{6^p}$
$y = \frac{1}{6^q}$
We also know that $x<y$, which effectively means $P>Q$

Also, $\frac{1}{x }+\frac{ 1}{y} = 6^p + 6^q = 7992.$
Knowing $P>Q$, we can say: $6^p + 6^q = 6^q (6^{p-q}+ 1) = 7992.$

$7992 = 6^3 * 37$

Obviously, then, $Q=3$, and $37 = 36 + 1 = 6^2 + 1$
So $P - Q = 2 $ and therefore $P = 2+ 3 = 5$

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Spent Ages on this Question: (need to learn probabilty again :'( )

P=5, Q=3 IMO

$\frac{1}{x}+\frac{1}{y}=7992$

given that Probability of agate appearing all P times = x
therefore, $x=\frac{1}{(6^p)}$

similarly $y=\frac{1}{(6^q)}$

since y>x, this means P>Q

setting up the equation again

$\frac{1}{\frac{1}{6^P}} + \frac{1}{\frac{1}{6^Q}} = 7992$

$6^P+6^Q=7992\\
6^Q(6^{P-Q} + 1) = 6^3 \times 37$

this means that $6^Q=6^3$ and $6^{P-Q}+1=37$

hence Q=3 and P=5

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10 Jul 2024, 22:45

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(1/6)^P=x

(1/6)^Q=y

x<y so P>Q

1/x+1/y=7992

Doing the sum

x+y/(xy)= 7992
substituting
(1/6)^P=x

(1/6)^Q=y

we have
((1/6^P)+(1/6^Q))/((1/6^P)*(1/6^Q))=7992
((6^Q+6^P)/(6^P+Q))/1/(6^P+Q)=7992
6^Q+6^P=7992
6^Q+6^P=(2^3)*(3^3)*37
6^Q+6^P= (6^3)* 37
6^Q*(1+6^(P-Q))=(6^3)* 37
so Q=3

1+6^(P-Q)= 37
6^(P-3)= 36
6^(P-3)= 6^2
so
P-3=2
P=5

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