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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero

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03 Jul 2025, 11:21

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a_5 = a_4*(a_3)^3

Need to get a_4 out of the picture to test options.

a_4 = a_3*(a_2)^3 = a_3 * 8

Replace:

a_5 = 8 * a_3 * a_3^3 = 8 * a_3^4

Now find a pair of options that work in this equation:

a_5 = 128
a_3 = -2

Note: There are more numbers satisfying the equation, that's why you have to look at the options available.

Bunuel

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The infinite sequence of nonzero integers $a_1$, $a_2$, $a_3$, ... is such that $a_n = (a_{n-1})(a_{n-2})^3$ for $n > 2$, and $a_2 = 2$.

Select for $a_3$ and $a_5$ the values of $a_3$ and $a_5$, respectively, that would be jointly consistent with the given information. Make only two selections, one in each column.

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03 Jul 2025, 11:56

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Ans: a3 = -2 and a5 = 128

The infinite sequence of nonzero integers $a_1$, $a_2$, $a_3$, ... is such that $a_n = (a_{n-1})(a_{n-2})^3$ for $n > 2$, and $a_2 = 2$.

Because we do not know the value of a1, which is crucial for getting to the series, we can start backwards with the answer choices.
I will write a1 as a only
So, lets start with last option -8
lets say a3 = -8 = a3 * a^3 ==> a^3 = -4 (not possible because sequence is of nonzero int)

a3 = -2 =means=> a1 = -1
from this
a4 = -2* 2^3 = -16
a5 = 128
These two are in the options. (answer: a3 = -2 and a5 = 128)

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03 Jul 2025, 12:22

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[ltr]as per the given information, we can write
a5 = (a4 )(a3)^3
and a4 = (a3)(a2)^3, and a2= 2
then a4 = 8a3
a5 = 8 (a3)^4
in given option only a3 = -2 and a5= 128 satisy the eqaution

then ans is a3 = -2 and a5 = 128[/ltr]

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03 Jul 2025, 12:40

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IMO Answer is :a3= -2 and a5= 128

Given an=(an−1)(an−2)^3 for n>2 and a2=2

a3= a2* a1^3= 2 (a1)^3

a4= a3* (a2)^3 = 2* (a1)^3 * 2^3= 16 (a1)^3

a5= a4 *(a3)^3= 16 (a1)^3 * (2* (a1)^3)^3= (16*8)* (a1)^12= 128 (a1)^12

so a3= 2 (a1)^3 and a5=128 (a1)^12

since (a1)^3 is raised to an odd power, (a1)^3 can have a negative value and also (a1)^12 is raised to an even power it will have a positive value and looking at the answer choices, a1=-1
Hence a3 =2*(-1)^3=-2 and a5= 128 *(-1)^12=128

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a3 = -2 (Option E) and a5 = 128 (Option A) are the correct answers.

First lets understand the information mentioned in the question and what we need to find in order to answer the question.

First of all the question tells us that the sequence is a infinite sequence of nonzero integers a(n) = (a(n-1))*(a(n-2))^3 for n>2 terms, and a2 = 2. And then the question asks us to find the value of a3 and a5.

Now lets put the value in the given formula and find out what values we will get.

a3 = (a(3-1))*(a(3-2))^3
[ltr]a3 = (a2)*(a1)^3
a3 = 2*(a1)^3

a4 = (a(4-1))*(a(4-2))^3
a4 = (a3)*(a2)^3, from the question we all know a2 = 2, so (a2)^3 = 8
a4 = 8*(a3)
a4 = 16*(a1)^3

a5 = (a(5-1))*(a(5-2))^3
a5 = (a4)*(a3)^3
a5 = (16*(a1)^3)*(2*(a1)^3)^3
a5 = (16*(a1)^3)*(8*(a1)^9)
a5 = 128*(a1)^12
[/ltr]
From here we know that the key to finding the values of a3 and a5 is to first find the value of a1. But now we don't have any addition information as we have already used all the information which was given in the question. So lets try finding the value of a1 by using the answer options.

So first lets try to find the value of a1 by putting the option values in a3 as it seems easier to solve than solving for a5.

As we call can check from the calculation which we have done till now the value of a5 will be greater than a3. So based on this calculation lets try checking the option from bottom.

First Option E: According to which a3 = -8
2*(a1)^3 = -8
(a1)^3 = -4
From here we can conclude that (-8) is not the right answer to the value of a3 because if we solve any further then a1 will be in decimal form which will defy the condition mention in the question i.e. all the number in series are nonzero integers. Eliminated

Now lets move on to next Option D: a3 = -2
2*(a1)^3 = -2
(a1)^3 = -1
(a1) = -1
This option fits all the condition mentioned in the question. So if we take the value of (a1) to be (-1) then from here a5 will be:
a5 = 128*(a1)^12
a5 = 128*1
a5 = 128 (Option A)

After solving this much we have finally got out answers which we were looking for. But lets check other options as well to be double sure.

These steps are just for verifying our option and that's why these are not necessary steps.

Option D: a3 = 4
2*(a1)^3 = 4
(a1)^3 = 2 Eliminated as we will get decimal value from here.

Option C : a3 = 8
(a1)^3 = 4 Eliminated as we will get decimal value from here.

Option B: a3 = 16
(a1)^3 = 8
a1 = 8
Now check for the value of (a5) = 128*(8)^12 and from here only we can understand the value which we will be getting is not even in the option so Eliminated.

Option A: a3 = 128
(a1)^3 = 64
a1 = 4
Again no need to calculate it further here also the value of (a5) = 128*(4)^12 is not in the option so Eliminated.

From here we can finally conclude that a3 = -2 (Option E) and a5 = 128 (Option A) are the only solution available.

Bunuel

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Given: $a_2$ = 2

Let's find the following:

1) $a_3$ = $a_2$* $a_1^3$ = 2 $a_1^3$-------1

2) $a_4$ = $a_3$ * $a_2^3$ = $a_2^4$ * $a_1^3$ = 16 $a_1^3$------2

3) $a_5$ = $a_4$ * $a_3^3$

On substituting the values from 1 and 2
$a_5$ = 128* $a_1$^12-------3

Now lets draw relation between $a_3$ , $a_5$ from 1 and 3 on solving

==> $a_5$ = 8 $a_3^4$

Lets check with options

If $a_3$ = -2
==> $a_5$ = 8 $a_3^4$ = 128

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Lets back solve this using options:

a5 can have a max value of 128.

So a3 cannot be 4, 8, 16, since cube of these numbers multiplied by a4 will be bigger than 128. It can also not be -8, since a number lesser that (-128)^3 is not in options.
a3 can be 4 or -2

We know that a2=2. Lets take a3 = -2.
a4 = -16
a5= 128[present in options]. Therefore is the correct answer.

Bunuel

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This is a tricky question.

First we need to understand that even though $a_2 = 2$ is given. we cannot find values without either $a_1$ or $a_3$. So first lets check with
$a_1$ = 1.
This will give you $a_3$ = $ 2* (a_1)^3 $ = 2 but 2 is not there in option. What if $a_1 = -1 $? then we have -2 as option.
Then we get

$a_4 = -2*2^3 = -16 $
and
$a_5 = -16*(-2)^3 = -16*(-8) = 128$.

So choose $a_3 = -2$ and $a_5 = 128.$

Bunuel

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Based on the given relation: $a_n = (a_{n-1})(a_{n-2})^3$ for $n > 2$, and $a_2 = 2$

=> $a_4 = (a_3)(a_2)^3 = (a_3)(2)^3 = 8(a_3)$

And, $a_5 = (a_4)(a_3)^3$

Substituting $a_4 = 8(a_3)$ from earlier,

=>$a_5 = 8(a_3)(a_3)^3$
=>$a_5 = 8(a_3)^4$

Plugging in $a_3 = -2$ will give us $a_5 = 8(-2)^4 = 128$

We can observe that all other values for $a_3$ will yield positive values for $a_5 > 128$, which are not available amongst the answer choices for $a_5$

Thus, the answer is $a_3 = -2$ and $a_5 = 128$

Bunuel

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Given:
a1 = a1, a2 = 2

Strategy: Find a5 and compare scenarios to answers
1. a5
a3 = (a2)*(a1)^3 = (2)*(a1)^3
a4 = (a3)*(a2)^3 = ((2)*(a1)^3) * (2)^3
= (16)*(a1)^3
a5 = (a4)*(a3)^3 = ((16)*(a1)^3) * ((2)*(a1)^3))^3
= (128)*(a1)^12
2. Scenarios
i) a1 = 1: a3 = (2)*(1)^3 = 2; a5 = (128)*(1)^12 = 128
ii) a1 = -1: a3 = (2)*(-1)^3 = -2, a5 = (128)*(-1)^12 = 128
3. Answers
Since 2 is an option and -2 is not, Scenario ii) holds

Answer: -2; 128

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Call the first term a1 just “k.”

The rule says each new term equals the previous term times the cube of the one before that.

So the third term works out to 2×k3.

That third term must be one of the choices given. The only way 2×k3 matches one of those choices is when k = –1 (because 2×(–1)3=–2).

Using k = –1 in the same rule again, the fifth term becomes 128×k12. Any number to the 12th power is positive, so this simplifies to 128.

Both answers (–2 for a3 and 128 for a5) are on the list, and no other starting value fits the criteria .

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Bunuel

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a2 = 2

a3 = a2 * a1^3

a4 = a3 * a2^3

= 8 * a3

a5 = a4 * a3^3

= 8 * a3 * 8 * a1^3

= 8 * 2 * a1^3 * 8 * a1^3

= 128 ^ a1^6

Now if a1 = -1

a^3 = -2
a^5 = 128

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a2= 2
a3= a2*a1= 2*(a1)^3

a4=(a3)*(a2)^3= (2*(a1)^3)*(2)^3= (2*(a1)^3)*(8)

a5=a4*(a3^3)= (2*(a1)^3)*(8) * (2*(a1)^3)^3
a5= 16*8*(a1)^12
a5=128*(a1^12)

therefore, a5 must be multiple of 128
and with the help of given options to select from we can choose -2,128 as our answers

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We can start building $a_5$ as follow:

$a_4 = (a_{3})(a_{2})^3 $
$a_5 = (a_{4})(a_{3})^3 $

Since $a_2=2 $, we can rewrite the equations:

$a_4 = 8*(a_{3}) $
$a_5 = (a_{4})(a_{3})^3 $

Then replace $a_4 $in the second equation with $ (a_{3})*8 $

$a_5 = 8*(a_{3})(a_{3})^3 = 8*(a_{3})^4 $

The only 2 options that satisfy the equation $a_5 = 8*(a_{3})^4 $ are :

$a_3=-2 $
$a_5=128 $

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The infinite sequence of nonzero integers a1 a 1 , a2 a 2 , a3 a 3 , ... is such that an=(an−1)(an−2)3 a n = ( a n − 1 ) ( a n − 2 ) 3 for n>2 n > 2 , and a2=2 a 2 = 2 . Select for a3 a 3 and a5 a 5 the values of a3 a 3 and a5 a 5 , respectively, that would be jointly consistent with the given information. Make only two selections, one in each column.

a3 = a2 x (a1)3 => a3/2 = (a1)3 => so first answer choice divided by 2 , needs to be an integer to the 3rd power => either 16/2 = 8 or -2/2 = -1

if we proceed with 16 as a3 to calculate a4 => we will arrive at 128 => so eliminate

therefore a3 has to be -2
A4 = -2 x (2)3 = -16
a5 = -16 x (-2)3 = 128

a3 = -2
a5 = 128

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For this problem I used the "pick numbers" strategy. Since a number is dependent of the two previous numbers, it will be an iterative method. To find a3 and a5 we need a1, a2, a3, a4, and a5.

a1 = ?
a2 = 2
a3 = 2
a4 = ?
a5 = ?

For n > 2, an = (a{n-1})*(a{n-2})^3

The options choices give us a sense that a1 could be positive or negative.

Trying negative, we can set a1 as -1.

Then, a3 = (2)*(-1)^3 = 2*(-1) = -2. Ok, that's a valid option choice for us. Lets continue.

a4 = (-2)*(2)^3 = (-2)*(8) = -16.
a5 = (-16)*(-2)^3 = (-16)*(-8) = 128. Ok, that's also a valid option choice for us!

Final answer: a3 = -2, a5 = 128.

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Bunuel

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See, we want ti find a5 and a3.

according to relation we can write a5 as ;

= a4 . (a3)cube
= (a3 . (a2)cube) . (a3)cube

So, a5 = ((a3)power4) . ((a2)power3)

Now, we can just put some value from the options for A3 (we already know A2) , WE CAN THEN GET RESULT FOR a5

So, when we put a3 = -2, we get a5 = 128. which is correct answer

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03 Jul 2025, 20:42

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a4 = a3 x a2^3 = 8a3
a5 = a4 x a3^3 = 8a3^4

a3 must be -2 and a5 must be 128. If a3 is bigger, the number a5 would be larger than available options.

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Correct Answer: a3= -2 & a5= 128
We need to assume a1 to get other values
Assume a1= -1
So a2= 2 (Given)
a3= (an-1) (an-2)^3
(2) (-1)= -2
a4= (an-1) (an-2)^3
(-2) (8)= -16
a5= (an-1) (an-2)^3
(-16) (-8)= 128

We can also put a1= 1 but we will get a3= 2 but option have -2 so well will put a1= -1.

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From the sequence
a_4 = a_3 * (a_2)^3 = a_3 * (2^3) = 8*a_3
a_5 = a_4 * (a_3)^3 = 8*a_3*(a_3)^3 = 8*(a_3)^4
substitute options
if a_5 = 128 then a_3^4 = 128/8 = 16
hence a_3 = +2 or -2
hence a_3 = -2 as +2 isn't there in options

Bunuel

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