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

1 2 3 4 5 6

AA155

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03 Jul 2025, 21:12

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Ans. a5 = 128 and a3 = -2

let a1=x a2=2

a3= a2.x^3 = 2x^3

a 4= a3.a2^3= 8.a3

a5= a4.a3= 8a3.a3^3= 8a3^4

use answer choices to evaluate; you will get x = -1.

i.e., a3 = -2
a5= 8. -2^5= 128

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, 21:13

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We have a5= a4 * a3^3 => a4=a5/a3^3 (1)
a4= a3 * a2^3= a3 * 2^3 = 8*a3 (2)
From (1) and (2), a5/a3^3 = 8 *a3 => a5= 8*a3^2. Use the answer in the table can easily find that a3=4 and a5 =128

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03 Jul 2025, 21:19

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$a_n=(a{n−1})(a_{n−2})^3$ & $a_2=2$

$a_5=a_4*a_3^3$

$a_4=a_3*a_2^3=a_3*2^3=8a_3$

Then, $a_5=8a_3*a_3^3=8a_3^4$

$a_5>a_3$.

When substitute $-2$ for $a_3$

$a_5=8*(-2)^4=8*16=128$

$a_3:$ -2
$a_5:$ 128

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03 Jul 2025, 21:44

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Bunuel

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$a_3$ = $a_2$*($a_1$)^3 = 2* ($a_1$)^3
$a_4$ = $a_3$*($a_2$)^3 = 2* ($a_1$)^3 * 2^3 = 16* ($a_1$)^3
$a_5$ = $a_4$*( $a_3$)^3 = 16* ($a_1$)^3 * 2^3* ($a_1$)^9 = 128* ($a_1$)^12

Try $a_1$= 1
=> $a_3$ =2 and $a_5$=128 (there is no option for -2 but yes option for 128)
Try $a_1$= -1
=> $a_3$= -2 and $a_5$=128 THIS IS THE RIGHT ANSWER!

naman.goswami

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03 Jul 2025, 22:08

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a3 = (a2)(a1^3)
a3= 2 (a1^3)-----------------------------------eq1

a4= a3(a2^3)= 8 a3

a5= (a4)(a3^3)= (8a3)(a3^3) = 8a3^4---------------------eq2

With this information let's try putting Different values of a3 and check if any of those satisfies eq 2 as well for a5

If a3 = 128
a5 = 8(128^4) does not satisfy any option

If a3= 16
a5 = 8(16^4) does not satisfy any option

if a3 = 8 , a5 = 8(8^4)does not satisfy any option
if a3= -2 , a5 = 8(-2^4) = 128 satisfies the option hence it is the answer

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03 Jul 2025, 22:08

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we are given a recurrence:
an = (an−1)(an−2)^3 for n > 2
we're also given a2 = 2
we are not directly told a1, so we’ll test the available answer choices for a3 and a5 and work backwards

start by testing a3 options
try a3 = 8
then a3 = (a2)(a1)^3
8 = (2)(a1)^3
so (a1)^3 = 4 → a1 = cube root of 4, which is not an integer → eliminate

try a3 = 16
16 = (2)(a1)^3
(a1)^3 = 8 → a1 = 2
this is valid
so if a1 = 2, a2 = 2, a3 = 16
compute a4 = (a3)(a2)^3 = 16 × 8 = 128
a5 = (a4)(a3)^3 = 128 × (16)^3
16^3 = 4096
a5 = 128 × 4096 = 524288
not among options → eliminate

try a3 = 4
4 = 2 × (a1)^3 → (a1)^3 = 2 → not an integer → eliminate

try a3 = -8
-8 = 2 × (a1)^3 → (a1)^3 = -4 → a1 not integer → eliminate

try a3 = -2
-2 = 2 × (a1)^3 → (a1)^3 = -1 → a1 = -1
possible
a1 = -1, a2 = 2, a3 = -2
a4 = a3 × (a2)^3 = -2 × 8 = -16
a5 = a4 × (a3)^3 = -16 × (−2)^3 = -16 × (−8) = 128
a5 = 128 is in the list

final selections:
a3 = -2
a5 = 128

FrontlineCulture

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03 Jul 2025, 22:37

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if a2 = 2
and a2= (a1)(a0) --> 2 is prime, so only factors are 2 and 1, and 2 to the third power = 8, so a1 must equal 2, and a0 must equal 1 to the 3rd power which equals 1.

therefore a3= (a2)(a1)^3 = (2)(2^3) = 16

got lost in the sequence for a5, but 128 seems like logical answer lol.

Bunuel

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03 Jul 2025, 22:41

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a3 = a2 . a1^3 = 2 . (a1 ^ 3)
a4 = a3. a2^3 = 2. (a1^3) . 8 = 16. (a1 ^3)
a5 = a4. a3^3 = 16 (a1 ^3) . [2. a1 ^3]^3 = 16. (a1^3) . 8 . a1^9 = 128. a1^12

so possible ans
a3 = 2; a5= 128
a3 = -2; a5 = 128
... (many more combinations but look at the choices avail and discard)

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03 Jul 2025, 22:44

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Main trick in this one was to figure out the value of a1 (not given).

By solving a3 = 2 a1^3
And a5= 2^7 a1^12

Now by looking at the options - Only consistent value satisfying bot would be when a1 is -1
Hence a3 = -2 and a5 = 128

Hope this is correct!

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03 Jul 2025, 23:04

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We are given the recurrence relation:
an = an-1 × (an-3)^3 for n > 2, and a2 = 2
Assume a1 = -1 ,Then:
a3 = a2 × (a1)^3 = 2 × (-1)^3 = -2
a4= a3× (a2)^3= (-2) × (2)^3 = -2 × 8 = -16
a5 = a4 × (a3)^3 = (-16) × (-2)^3 = -16 × (-8) = 128
So, a3 = -2 and a5= 128

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03 Jul 2025, 23:29

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Bunuel

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$a_2$ = 2
Let $a_1$ = x

From $a_3$ onwards we have
$a_n = (a_{n-1})(a_{n-2})^3$

=> $a_3$ = $a_2$ * ${a_1} ^3$ = 2 * $x^3$
=> $a_4$ = $a_3$ * ${a_2} ^3$ = 2 * $x^3$ * $2^3$ = 16 * $x^3$
=> $a_5$ = $a_4$ * ${a_3} ^3$ = 16 * $x^3$ * $2^3 * {x^3} ^3$ = 128 * $x^{12}$

Now considering $a_5$ = 128 * $x^{12} $ we know the highest number in the option is 128 hence we start by putting the value of 'x' = 1 or -1

=> $a_5$ = 128 * $1^{12} $ = 128

Now put the same value of 'x' in $a_3$ we get
$a_3$ = 2 * $1^3$ = 2
However, 2 is not in the options so x cannot be 1

Now using x = -1 we get
=> $a_5$ = 128 * ${-1}^{12} $ = 128
Now put the same value of 'x' in $a_3$ we get
$a_3$ = 2 * $ {-1} ^3$ = -2
This works

Hence the choices are
$a_3$ = -2
$a_5$ = 128

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03 Jul 2025, 23:29

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Let $a_1=x$ (non-zero integer)
$a_2=2$
$a_3=a_2\times a_1^3=2x^3$
$a_4=a_3\times a_2^3=16x^3$
$a_5=a_4\times a_3^3=128x^{12}$

Now, test values for $a_3$and $a_5$

For $a_3=4$,
$2x^3=4 \implies x^3=2 \to$ non-integer

For $a_3=8$,
$2x^3=8 \implies x^3=4 \to$ non-integer

For $a_3=16$,
$2x^3=16 \implies x^3=8\implies x=2$
Then $a_5=128x^{12} \implies128\times2^{12} \to$ too large, not in choices

For $a_3=-8$,
$2x^3=-8 \implies x^3=-4 \to$ no integer

For $a_3=-2$,
$2x^3=-2 \implies x^3=-1\implies x=-1$
Then $a_5=128x^{12} \implies128\times-1^{12} = 128 \to$Valid Choices

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

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04 Jul 2025, 00:07

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

a4 = (a3) (a2)^3 = 8(a3)

a5 = a4 * (a3)^3

Subsituting values:

a3 = 2(a1)^3 = 128 => a1 = 4, But a5 having (a3)^3 the value will be too high

a3 = 16 => a5 too large

Considering a possible solution for a3
a3 = -2 => a4 = -16 => a5 = -16 * (-2)^3 = -16 * -8 = 128

Therefore a3 = -2
a5 = 128

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04 Jul 2025, 00:11

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a2 = 2
a3 = a2*a1^3= 2*a1^3
a4 = a3*a2^3 = 8a3 = 16*a1^3
a5 = a4*a3^3= (16*a1^3)*(2*a1^3)^3= 128*a1^12

maximum posssible value given is 128, therefore a1 can either be 1 or -1

Bunuel

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04 Jul 2025, 01:01

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an=(an−1)(an−2)^3
=> a3 = 2*(a1)^3
=> a4 = 2*(a1)^3 * 8
=> a5 = 16*(a1)^3 * 8*(a1)^9

Let's assume, a1 = -1; => a3 = -2 ; a5 = 128.

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04 Jul 2025, 02:28

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

It is better to choose little numbers for a3 because, as the sequence has a power of 3, the following terms grow quickly.
If a1=-1, then a3=-2

a4 = -2*2^3 = -16
a5 = -16 * (-2)^3 = 128

a3=-2 and a5=128

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04 Jul 2025, 02:40

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Let a1 = k and a2 = 2
Based upon the info,
a3 = a2 * (a1)^3 = 2 * k^3
a4 = a3 * (a2)^3 = (2 * k^3) * 2^3 = 16 * k^3
a5 = a4 * (a3)^3 = (16 * k^3) * (2 * k^3)^3 -> 16 * k^3 * 8 * k^9 = 128 * k^12

We need a3 and a5 and k should be a nonzero integer. Let's try a few values for k
If k = 1
then a3 = 2 * 1^3 = 2 (not in options)
If k = –1
then a3 = 2 * (–1)^3 = –2
and a5 = 128 * (–1)^12 = 128

Answer: a3 = –2 ; a5 = 128

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04 Jul 2025, 03:18

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Bunuel

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a5 = a4 * a3^3
a4 = a3 * a2^3 = 8 * a3
=> a5 = 8 * a3^4
So a5 is a very large number compared to a3 and is always positive, so we try the number with the smallest a3 = -2 => a5 = 128
Final answer: a3 = -2; a5 = 128

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04 Jul 2025, 03:31

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Given = $(a_{n−1})*(a_{n−2})^3; a_2 = 2 $

We are to find $a_3$ & $ a_5$,
From the relation we have,

$a_4 = (a_{4-1})*(a_{4-2})^3$

$a_4 = (a_3)*(a_2)^3$; Substituting $a_2 = 2 $ we get,
$a_4 = 8(a_3)$

$a_5 = (a_{5-1})*(a_{5-2})^3$
$a_5 = (a_4)*(a_3)^3$
$a_5 = 8(a_3)*(a_3)^3$
$a_5 = 8(a_3)^4 $;
This gives us an equation which can be used to find the required answers from the selections(Trial and error)

Giving smaller values from the selection to $a_3$; $a_3 = -2$, Then as per the equation we have $a_5 = 8(-2)^4 = 128$ and this is consistent with the selections given.

Therefore, the correct answers are $a_3 = -2$ & $a_5 = 128$

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04 Jul 2025, 04:06

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Since we don't have the value of a1, let it be x

a3 = (a2)*(a1)^3 = (2)*(x)^3
a4 = (a3)*(a2)^3 = (2x^3)*(2)^3 = (16)*(x)^3
a5 = (a4)*(a3)^3 = 128*x^12

a3=2* x^3
a5=128* x^12

Matching with given options for various values of x:
At x = -1
Answers a3 = -2 , a5 = 128