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In a certain sequence of 10 positive numbers, each number after the fi

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niteshwaghray
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In a certain sequence of 10 positive numbers, each number after the fi [#permalink] New post   29 May 2017, 10:16
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In a certain sequence of 10 positive numbers, each number after the first is square root of the previous number. If the first number is 9, how many numbers are less than 1?


A. 0
B. 1
C. 2
D. 3
E. 4
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Re: In a certain sequence of 10 positive numbers, each number after the fi [#permalink] New post   29 May 2017, 12:39
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niteshwaghray wrote:
In a certain sequence of 10 positive numbers, each number after the first is square root of the previous number. If the first number is 9, how many numbers are less than 1?


A. 0
B. 1
C. 2
D. 3
E. 4


Here is a little trick: any positive integer root from a number more than 1 will be more than 1. So, it does not matter how many times we take the square root from 9, we'll still won't get a number less than 1. Meaning: \(\sqrt[1000000]{9}>1\).

Answer: A.
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Re: In a certain sequence of 10 positive numbers, each number after the fi [#permalink] New post   29 May 2017, 23:08
So in this sequence we are going to start from 9 and take square roots.. so square root of 9 which is 3, .. then square root of 3 which is approx. 1.732, .. then sq root of 1.732 .. and so on...

Its tricky but interesting - as long as a number is > 1, its square root will also be > 1... so in this list no matter how many times we take the square root - ALL the numbers that we end up with will always be > 1 (though the numbers will keep on decreasing)

Hence all numbers are > 1. None of the numbers is < 1.

Thus option A
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Re: In a certain sequence of 10 positive numbers, each number after the fi [#permalink] New post   01 Jun 2017, 10:25
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niteshwaghray wrote:
In a certain sequence of 10 positive numbers, each number after the first is square root of the previous number. If the first number is 9, how many numbers are less than 1?


A. 0
B. 1
C. 2
D. 3
E. 4


We note the following fact: If x is greater than 1, then the square root of x is also greater than one. This is because the product of two numbers between 0 and 1 is again between 0 and 1; if the square root of x was less than 1, then x would have to be less than 1 as well.

Now we can conclude that all the numbers in the sequence are greater than 1, since the numbers in the sequence are obtained by taking square roots of numbers that are greater than 1. Thus, none of the numbers in the sequence are less than 1.

Answer: A
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Re: In a certain sequence of 10 positive numbers, each number after the fi [#permalink] New post   18 Dec 2019, 10:08
niteshwaghray wrote:
In a certain sequence of 10 positive numbers, each number after the first is square root of the previous number. If the first number is 9, how many numbers are less than 1?


A. 0
B. 1
C. 2
D. 3
E. 4


Square of any positive number less than 1 is always less than 1, therefore all numbers in the series 9,3,1.7,.... will be greater than 1.

IMO A

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Re: In a certain sequence of 10 positive numbers, each number after the fi [#permalink] New post   23 Dec 2019, 18:24
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niteshwaghray wrote:
In a certain sequence of 10 positive numbers, each number after the first is square root of the previous number. If the first number is 9, how many numbers are less than 1?


A. 0
B. 1
C. 2
D. 3
E. 4


We must remember the rule that when we take the square root of a number greater than 1, the result will be always greater than 1. Therefore, although the first number is 9, regardless of how many square roots we take, there will not be any results less than 1.

Answer: A
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Re: In a certain sequence of 10 positive numbers, each number after the fi [#permalink] New post   23 Feb 2024, 12:17
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