Latest Elitmus Aptitude Question SOLUTION: y=x^4+4 x=any 5digit no.. find probability that y is divisible by 5??

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(#M40154586) ELITMUS Numerical Ability question Keep an EYE

y=x^4+4
x=any 5digit no..
find probability that y is divisible by 5??

Asked In Elitmus ananga (8 years ago)

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Dheeraj Kumar  (8 years ago)

N can be any number so it will be 0.1,2,...9
x=-----(5 digit no)
y=(x^4+4)/5;
so the value of x can be as a last digit 1,2,3,4,6,7,8,9
total possible value is (0 to 9)=10
so probability is =8/10=4/5
i.e.
(0)^4+4=4(not divisible)
(1)^4+4=5
(2)^4+4=20;
(3)^4+4=85;
(4)^4+4=260;
(5)^4+4=___9(not divisible)
(6)^4+4=__0;
(7)^4+4=__5;
(8)^4+4=__0;
(9)^4+4=__5;

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Devulapalli Surender  (8 years ago)

Number divisible by 5 when last digit ends with either 0 or 5
Last digit becomes 5 when x^4 ends with 1
Last digit becomes 0 when x^4 ends with 6
probability of 1 or 6 =>2/10=> 1/5

Is this solution Helpfull? Yes (7) | No (10)  

Mithilesh Times  (8 years ago)

4/5 is correct ans

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JANARTHANAN RAM  (8 years ago)

Ans: 4/5

Total possible 5 digit numbers=90000
y=(x^4+4) & y%5=0
So unit digit must be 1 or 2,3,4,6,7,8,9.
Possible numbers=9*10*10*10*8=72000
So probability is =72000/90000=4/5

Is this solution Helpfull? Yes (5) | No   

Govind kumar  (8 years ago)

N is natural number so it will be 1,2,...9
x=-----(5 digit no)
y=(x^4+4)/5;
so the value of x can be as a last digit 1,2,3,4,6,7,8,9
total possible value is (1 to 9)=9
so probability is =8/9;
i.e (1)^4+4=5
(2)^4+4=20;
(3)^4+4=85;
(4)^4+4=260;
(6)^4+4=__0;
(7)^4+4=__5;
(8)^4+4=__0;
(9)^4+4=__5;
prob=8/9;

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sachin kumar  (8 years ago)

answer : the five digit no. X will be divisible only when it has 5 "or" 0 in the one's position .. and the a five digit no. is possible when the one's position is 0, 1, 2, 3, 4 ,5 ,6 ,7 , 8, 9 ..i.e, in 10 ways ..but for divisibility from 5 there are 5 cases possible when in these digits are in one's position 1=> 1+4=5 , 3=>1+4=5,7=>1+4=5,9=>1+4=5
So, 4/10 = 2/5 Answer

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