Latest Elitmus Aptitude Question SOLUTION: [n^1/3]=5.how many values are exist for n?

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(#M40025825) ELITMUS question Keep an EYE

[n^1/3]=5.how many values are exist for n?

Asked In Elitmus vamsikrishna kaluvakolanu (11 years ago)

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navneet kumar verma  (11 years ago)

ans will be 91.
explanation:
n^1/3 can range from 5.0 to 5.9 then the expression [n^1/3]=5..i.e greatest integer concept.
so now, we know:
5^3=125=>n [125^1/3]=[5]=5
6^3=216=>n [216^1/3]=[6]=6
so the value of n can range from [125,215] because if n=216 [n^1/3] will be 6.
hence total values for n=215-125+1=91 //+1 for including both boundary values 125 and 216.

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rajveer singh  (11 years ago)

hi vk take [ ] this as window for which there are number of values are available. since 5^3 is 125 and 6^3 is 216 so in between them 90 values are possible.

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vamsikrishna kaluvakolanu  (11 years ago)

options-
90
91
1
3

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gyaneshvari  (11 years ago)

[n^1/3]=5 as we know 125^1/3 =5 and 216^1/3= 6.....
so to satisfy above eqation we can take values of n from 125 to 215 both inclusive..[n] where n is greatest integer less than or equal to n.so total values psbl is 125-215+1=91..

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shipra jain  (11 years ago)

only one value for n=125
n^1/3=5
(n^1/3)^3=(5)^3
n=125

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naga sarath chand  (11 years ago)

explain me clearly,they have given that n^1/3 = 5 then how it lies between 5 and 6..no meaning

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Alok Gayakwad  (11 years ago)

only 1 value will satisfy for n^1/3=5, ie- n=125;
n^1/3=5
(n^1/3)^3=(5)^3 // will find ^3 for equation
n=125

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