Latest Elitmus Aptitude Question SOLUTION: If N = 82pow3 - 62pow3 - 20pow3 then N is divisible by: a) 31 and 41 b) 13 and 67 c) 17 and 7 d) none

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(#M40155594) ELITMUS Numerical Ability question Keep an EYE

If N = 82pow3 - 62pow3 -
20pow3 then N is divisible by:
a) 31 and 41
b) 13 and 67
c) 17 and 7
d) none

Asked In Elitmus Rizul (8 years ago)

Unsolved Read Solution (6)

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RAKESH  (8 years ago)

a) 31 and 41

value of N is = 82^3 - 62^3 -20^3 = 3*82*62*20 [a+b+c=0 => a^3+b^3+c^3=3abc]

= 3*(2*41)*(2*31)*20 divisible by both 41 & 31.

Is this solution Helpfull? Yes (36) | No (1)  

thjkhffn  (8 years ago)

L.C.M of ( 82, 62, 20) = 20*41*31
now,
from take the LCM of options one by one
i.e, a) 31*41 b) 13*67 c) 17*7
now , divide 20*41*31 by options
so you will get option (A) that is dividing perfectly

Is this solution Helpfull? Yes (7) | No   

ajnabee  (8 years ago)

N = 2^3 ( 41^3 - 31^3 - 10^3 )
N = 8 ( ( 41-31 ) ( 41^2 + 31^2 + 41*31 ) - 10^3 )
N = 80 ( 1681 + 961 + 1271 - 100 )
N = 80 ( 3813 )
N = 80 ( 41 ) ( 93 )
N = 80 "(41) (31)" 3
Option a)

Is this solution Helpfull? Yes (5) | No   

AVINASH PANDEY  (8 years ago)

STEP1
Take 2^3 common from the given equation
Hence, we'll get as:-
2^3 {(41)^3 - (31)^3 - (10)^3}

Step 2:-
Now since from the basic arithmetic we know that : a^3 - b^3 = (a-b)*{(a)^2 + (b)^2 + a*b)}
Hence just for the sake of ease of the following question I am considering: a=41 & b=31.
So, we'll get 2^3* [(a-b)*{(a)^2 + (b)^2 + ab}]

Step 3.
Substitute the value accordingly and by solving it further we'll get the following:
2^3*[{41-31}*{(41)^2 + (31)^2 + 41*31} - 10^3] which will in turn become (after taking 10 as common) as,
8*10[{a^2 + b^2 + ab} - 10^2]

Step 4.
8*10[{a^2 + b^2 + 2ab - ab} - 10^2]
80[{a + b}^2 - ab - 10^2 }]

Step 5.
Now finally substitute the values of a and b as 41 and 31 respectively .
Finally you will land upon the equation as follows:
2^3 *10 [ {(41)^2 + (31)^2 + 2*41*31} - {41*31} - {10}^2]
8*10[{41 + 31}^2 - 10^2 - 41*31]
80[ {62 * 82} - {41 * 31}]
80[ 2{31 * 41} - {41*31}]
80 * (41 * 31) [2 - 1]
80*(41*31)
this will be the outcome of the initial equation which is divisible by both 31 & 41.

HOPE YOU GOT IT :)

Is this solution Helpfull? Yes (2) | No (4)  

Prayosha panchal  (8 years ago)

a). 31 and 41

82^3-(62^3)-(20^3) => We can simplify this as

(62=20)^3- (62^3)-(20^3)
We know the formula that- (a + b)3 = a3 +b3 +3ab(a +b)
=>(a + b)3 -( a3 +b3) = 3ab(a +b)
that is 3*20*62*(20+62)
=3*20*62*82


by seeing we can say that is devisible by 31 and 41. As it has 62 and 82 in the multiplication.

Hope this is simpler to understand.

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pankaj yadav  (7 years ago)

a) 31 and 41

Value of N is = 82^3 - 62^3 -20^3 = 3*82*(-62)*(-20)
because
if a+b+c=0 then a^3+b^3+c^3=3abc

= 3*(2*41)*(2*31)*20 divisible by both 41 & 31.

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