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655-705 Level|   Math Related|         
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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 09:47
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a3 = a2 * (a1)^3 = 2 (a1)^3
a4 = a3 * (a2)^3 = 16 (a1)^3
a5 = a4 * (a3)^3 = 128 (a1)^12

With the given options and equations above, we can observe that a3 < a5
Also a3 can be negative but a5 is always positive
and multiplication factor is 128 for a5

So substituting the low value for a3 trying with -2
a1 = -1
a5 = 128

Hence this satisfies
We do not find any combination apart from this and is thus correctanswer

a3 = -2
a5 = 128

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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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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 09:48
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okay so

a3 = 2 * a1^3
a4 = 2 * a1^3 * a2^3 = 16 * a1^3
a5 = a4 * a3^3 = 16*a1^3 * 8*a1^9 = 128*a^12

Now if a1 = 1 then ans a3 = 2, a5 = 128
if a1 = -1 then ans a3 = -2 , a5 = 128
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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 09:51
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a5 = a4 * (a3)^3;
a4 = a3 * (a2)^3;

Plug-in the value a3 = 4;
a4 = 4 * 8 = 32
a5 = 32 * 64; (Eliminate)

Plug-in the value a3 = -2;
a4 = -2 * 8 = -16
a5 = -16 * -8 = 128;

Therefore,
a3 = -2; Option E
a5 = 128; Option A
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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 09:53
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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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given a2 = 2

and a(n) = a(n-1)*(a(n-2))^3

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

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

and a5 = a4 * (a3) ^3 = 32 * (a1)^6, Since there is only one option that is a multiple of 32 Therefore a5= 128. Option A or 128

And from a5 = 128 = 32 * (a1)^6 we can get the value of a1 = 2^(1/3)

then the value of a3 = 2 *(2 ^(1/3)) ^3 = 4. Hence Option D or Option 4
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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 09:54
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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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Solved using hit and trial method of options
asuming a3=-8
-8 = 2 * a1 ^3 , so a1 = -1

and therefore a5 = a4 * a3^3 = 128

putting all other options as a3 produces non integer
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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 09:54
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from above equation
a3=a2*a1^3= 2a1^3

a4=a3*a2^3=8a3

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

lets take a1=2
a3=4
a5=8*16*16

not possible
a1=1
a3=2
a5=128
Not possible

a1=-1
a3=-2
a5=128
Correct answer is a3=-2 & a5=128
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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 10:01
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Made two equations basically:

a4=a3*2^3=8(a3)
a5=8(a3)*a3

Only a3=8 and subsequently a5=128 satisfy that equation. Hence these two should be the options
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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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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 10:02
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a3=(a2) X (a1)^3
a3= 2(a1)^3
Similarly,
a5=(a4) X (a3)^3
a5=((a3) X (a2)^3) X (2(a1)^3)^3
a5= (2(a1)^3 X 8) X 8(a1)^9
a5= 128(a1)^12
a5=(2^7) X (a1)^12
a5=(2^3) X [2(a1)^3]^4
now substituting a3 in this equation,
a5=8(a3).
So among the options,
only values for a3 and a5 that satisfy the above equation are 16 & 128 respectively.
So, a3=16 and a5=128
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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 10:02
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We are given the recurrence relation a(n) = a_(n-1) * (a_(n-2))^3 for n > 2, with a_2 = 2. The term a(1) must be a non-zero integer.

First, let's express the terms we need, a_3 and a_5, using the given information.

For n = 3, the formula gives:
a_3 = a_2 * a_1^3
Since a(2) = 2, this becomes:
a_3 = 2 * (a_1)^3

Next, let's derive an expression for a_5. We first need a_4.
For n = 4: a_4 = a_3 * (a_2)^3 = a_3 * (2)^3 = 8 * a_3.
For n = 5: a_5 = a_4 * a_3^3.
Substituting our expression for a_4, we get:
a_5 = (8 * a_3) * (a_3)^3 = 8 * (a_3)^4.

So, we have two key equations:
1. a_3 = 2 * (a_1)^3
2. a_5 = 8 * (a_3)^4

The first equation tells us that a_1 is the cube root of a_3/2. Since a_1 must be a non-zero integer, we can test the options provided for a_3 to see which one yields an integer value for a_1.

Let's test the option a_3 = -2.
If a_3 = -2, then (a_1)^3 = -2 / 2 = -1. The cube root of -1 is -1, so a_1 = -1. This is a valid non-zero integer.

Now, we use this value of a_3 to find the corresponding value of a_5 using our second equation.
If a_3 = -2, then a_5 = 8 * (-2)^4 = 8 * 16 = 128.

The pair of values a_3 = -2 and a_5 = 128 are both present in the options list and are consistent with the sequence's rules, originating from a_1 = -1. Checking other options for a_3 will not produce a value for a_5 that is also in the list.

Therefore, the correct value for a_3 is -2 and for a_5 is 128.

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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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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 10:05
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We know that \(a_2 = 2\). Basis the expression given, we can write:

\( a_3 = a_2 * a_1^3 = 2 * a_1^3\)

Similarly, we can write:

\(a_4 = a_3 * a_2^3 = 2^4 * a_1^3\)

\(a_5 = a_4 * a_3^3 = 2^7 * a_1^1^2\)


Looking at just the expression we got for \(a_3\) and \(a_5\),

We can say that if \(a_3 = m\), then \(a_5 = 8m^4\)

Now we just need to find a value for \(m\), that satisfies the values for both \(a_3\) and \(a_5\),

At, \(m = -2\), we get \(8m^4 = 128\)

Thus, these are the correct corresponding values.

\(a_3 = -2\)
\(a_5 = 128\)
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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 10:05
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We are told that each number in the sequence depends on the two numbers before it, and that the second number is 2. To find which values of the third and fifth numbers fit this rule, we can start by guessing a value for the first number. If we assume the first number is -1, then the third number comes out to be -2 using the given rule. Continuing the pattern using these values, the fifth number works out to be 128. This matches the choices in the table, so the correct pair that follows the rule is the third number as -2 and the fifth number as 128.

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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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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 10:06
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Given a2 = 2, using given equation, following values can be calculated in terms of a1

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

Substituting some random values of a1 such as 1,-1 and 1/2,

Only a1= -1, a3=-2 and a5 = 128 satisfies the above equations.
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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 10:07
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We will determine possible values for a3:
n=3,
a3=a2*(a1)^3=2*(a1)^3

We will look for form 2k^3 from answer choices:
valid a3 candidates are 128, 16, and −2.

Now, for a5:

a4=a3*(a2)^3=> a3*8
a5=a4*(a3)^3

From answer choices:

If a3=−2,
a4=−2*8=−16
a5=−16*(−2)^3
=−16*(−8)=> 128

Therefore, a3=-2 and a5=128.
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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 10:10
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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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We know that a5 = a4 (a3)^3 as per the given equation.

We also know that a4 = a3(a2)^3 => 8a3. So if we substitute this a5 = 8a3 (a3)^3 => a5 = 8(a3)^4.

So we need to find the matching pair \(a5 = 8(a3)^4\). If a3 = -2, a5 = 128.

Hence IMO a3 = -2 , a5 = 128.
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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 10:13
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let a1=x
a3=a2(a1)3=2.x3

a5=a4(a3)3=a3(a2)3(a3)3
=(a3)4*8=128*x12

let us assume a1=1,
a3=2, a5= 128--No
let us assume a1 as -1;
a3=-2, a5=128.

Hence this will be answer.
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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 10:41
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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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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 10:55
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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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We have to ensure the unknown value as x form and equation then use the options to solve it.

Given:
  • a2 = 2
  • Formula: an = a(n−1) * (a(n−2))^3 for n > 2
  • Let a1 = X
Then:
  • a3 = 2 * X^3
  • a4 = a3 * 2^3 = (2 * X^3) * 8 = 16 * X^3
  • a5 = a4 * (a3)^3 = (16 * X^3) * (2 * X^3)^3
    = 16 * X^3 * 8 * X^9 = 128 * X^12
X cant be a super big value as over the given the options maximum value for a5 is 128 so either must be +1/-1.
Now try values:

If X = -1:
  • a3 = 2 * (-1)^3 = -2
  • a5 = 128 * (-1)^12 = 128

Hence :
a3 = -2, a5 = 128
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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 11:11
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a2=2 => a3=2 x (a1)^3 => a3 should be a number that (a3/2) is a cubic of an integer (because a1 is an integer) => Only 16 and -2 is reasonable for being a3 (16/2=8=2^3 and -2/2=-1=(-1)^3)
If a3=16 => a4=16x2^3=16x8=128 => a5= 128x16^3 => So big for any other options to choose.
If a3=-2 => a4=(-2)x2^3=-16 => a5=(-16)x(-2)^3=1288 => a3=-2 and a5=128

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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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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 11:15
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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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The sequence is mentioned in the question stem for n>2

\(a_n = (a_{n-1})(a_{n-2})^3\) for \(n > 2\), and \(a_2 = 2\).

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 = 128 (a1)^12


a1 cannot be 1. Then the values of a3 and a5 becomes 2, 128.

if we take a1 = -1 ( non zero integers ) , then a3 = -2 and a5 = 128

Thus, the values of a3 = -2 , a5 = 128.
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GMAT Club Olympics 2025 (Day 4): The infinite sequence of nonzero 03 Jul 2025, 11:17
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Given :
an=(an−1)(an−2)^3 for n>2 and an = 2

Let’s assume a1=1 or -1

Assume a1=1:
  • a2=2 (given)
  • a3=2*(1)^3=2
  • a4=2*(2)^3=2*8=16
  • a5=16*(2)^3=16*8=128
So: a3 = 2, a5 = 128



Assume a1=-1:
  • a2=2 (given)
  • a3=2*(-1)^3=-2
  • a4=-2*(2)^3=-2*8=-16
  • a5=-16*(-2)^3=-16 (-8) =128
So: a3 = -2, a5 = 128


ans: a3 = -2, a5 = 128
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