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Number Systems -Sequence and Series

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by sukhman » Thu Oct 17, 2013 8:19 am
In a certain sequence, every term after the first is determined by multiplying the previous term by an integer constant greater than 1. If the fifth term of the sequence is less than 1000, what is the maximum number of nonnegative integer values possible for the first term?
A) 60 B) 61 C) 62 D) 63 E) 64
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by GMATGuruNY » Thu Oct 17, 2013 8:41 am
sukhman wrote:In a certain sequence, every term after the first is determined by multiplying the previous term by an integer constant greater than 1. If the fifth term of the sequence is less than 1000, what is the maximum number of nonnegative integer values possible for the first term?
A) 60 B) 61 C) 62 D) 63 E) 64
Let a� = the first term and a₅ = the fifth term.

To MAXIMIZE the number of options for a�, we must MINIMIZE the multiplier for the sequence.
Since the multiplier must be greater than 1, the smallest possible multiplier between successive terms = 2.
Implication:
a₅ = a� * 2 * 2 * 2 * 2
a₅ = 16a�.

Since aâ‚… < 1000, we get:
16a� < 1000.
a� < 1000/16.

Since 1000/16 ≈ 62, any integer between 0 and 62, inclusive, could be the value of a�:
If a� = 0, then a₅ = 16(0) = 0.
If a� = 1, then a₅ = 16(1) = 16.
If a� = 62, then a₅ = 16(62) = 992.
In every case, aâ‚… < 1000.

Since a� could be any integer between 0 and 62, inclusive, the total number of options for a� = 63.

The correct answer is D.
Last edited by GMATGuruNY on Mon Jul 23, 2018 6:43 pm, edited 1 time in total.
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by alanforde800Maximus » Mon Jul 23, 2018 6:26 pm
GMATGuruNY wrote:
sukhman wrote:In a certain sequence, every term after the first is determined by multiplying the previous term by an integer constant greater than 1. If the fifth term of the sequence is less than 1000, what is the maximum number of nonnegative integer values possible for the first term?
A) 60 B) 61 C) 62 D) 63 E) 64
Let a� = the first term and a₅ = the fifth term.

To MAXIMIZE the number of options for a�, we must MINIMIZE the multiplier for the sequence.
Since the multiplier must be greater than 1, the smallest possible multiplier between successive terms = 2.
Implication:
a₅ = a� * 2 * 2 * 2 * 2
a₅ = 16a�.

Since aâ‚… < 1000, we get:
16a� < 1000.
a� < 1000/16.

Since 1000/16 ≈ 62, any integer between 0 and 62, inclusive, could be the value of a�:
If a� = 0, then a₅ = 16(0) = 0.
If a� = 1, then a₅ = 16(1) = 1.
If a� = 62, then a₅ = 16(62) = 992.
In every case, aâ‚… < 1000.

Since a� could be any integer between 0 and 62, inclusive, the total number of options for a� = 63.

The correct answer is D.
Hello Mitch,

I guess you meant to write 16 here:-

If a� = 1, then a₅ = 16(1) = 1.
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by GMATGuruNY » Mon Jul 23, 2018 6:45 pm
alanforde800Maximus wrote:Hello Mitch,

I guess you meant to write 16 here:-

If a� = 1, then a₅ = 16(1) = 1.
Good catch!
I've corrected the typo.
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