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04 Apr 2020, 13:49
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This is how i solved..does this work?
Terms are multiplied by a constant integer.
Smallest integer greater than 1 is 2
62*2*2*2*2 = 62*16 = 992
63*2*2*2*2 = 63*16 = 1008 ---> too large
Range is from 0-62 = 63 numbers
Terms are multiplied by a constant integer.
Smallest integer greater than 1 is 2
62*2*2*2*2 = 62*16 = 992
63*2*2*2*2 = 63*16 = 1008 ---> too large
Range is from 0-62 = 63 numbers
vaibhav1221
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09 Jan 2022, 06:44
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Very nice question. There is a decent chance that you'd get this question wrong even after figuring out the logic.
Here is how I did it.
Assume
\(S1 = (n^0)x\) -> First Term
\(S2 = (n^1)x\)
\(S3 = (n^2)x\)
\(S4 = (n^3)x\)
\(S5 = (n^4)x\) -> 5th Term, which is less than \(1000\).
Let's say the \(S5\) is \(999\).
Possible values of \(n\)
\(n≥2\) for \(x\) ∈ [0,∞) and \(n>1\) ∈ Integers
Because the smallest value of n will render the maximum values of \(x\) as \(2^4x\) can have the maximum value for any \(x\) ∈ [0,∞) and \(2^4x ≤ 999\)
So, \(2^4x ≤ 999\)
\(16x ≤ 999\)
\(x ≤ \frac{999}{16}\)
\(x ≤ 62.5\)
So, \(x\) can take values up to the integer \(62\) for \(x\) ∈ [0,∞)
\(S1\) can take values - (0,1,2,3,4...62) 63 values. So, D is the correct answer choice.
Here is how I did it.
Assume
\(S1 = (n^0)x\) -> First Term
\(S2 = (n^1)x\)
\(S3 = (n^2)x\)
\(S4 = (n^3)x\)
\(S5 = (n^4)x\) -> 5th Term, which is less than \(1000\).
Let's say the \(S5\) is \(999\).
Possible values of \(n\)
\(n≥2\) for \(x\) ∈ [0,∞) and \(n>1\) ∈ Integers
Because the smallest value of n will render the maximum values of \(x\) as \(2^4x\) can have the maximum value for any \(x\) ∈ [0,∞) and \(2^4x ≤ 999\)
So, \(2^4x ≤ 999\)
\(16x ≤ 999\)
\(x ≤ \frac{999}{16}\)
\(x ≤ 62.5\)
So, \(x\) can take values up to the integer \(62\) for \(x\) ∈ [0,∞)
\(S1\) can take values - (0,1,2,3,4...62) 63 values. So, D is the correct answer choice.
22 Oct 2025, 06:39
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how do you conclude that each time we are multiplying the previous term with an integer, we are multiplying with the same integer every time?
Bunuel
