Latest Elitmus Aptitude Question SOLUTION: what is the probability of dividing a randomly selected number (x^4 + 1) by 5,where as x is a four digit number?

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(#M40017894) ELITMUS Numerical Ability question divisiblty Keep an EYE

what is the probability of dividing a randomly selected number (x^4 + 1) by 5,where as x is a four digit number?

Asked In Elitmus sumit (11 years ago)

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abhishek kumar  (11 years ago)

(x^4 + 1) is divisible by 5 when last digit is 0 or 5.
if we replace X by any of d no: from 0 to 9, non of d no: contain last digit 0 or 5.
hence probability is 0.

Is this solution Helpfull? Yes (42) | No (7)  

akhilesh kumar  (11 years ago)

unit place of x could be 0-9;
unit place of x^4 will be any of these
0^4=0;1^4=1;2^4=6;3^4=1;4^4=6;5^4=5;6^4=6;7^4=1;8^4=6;9^4=1
unit plase of x^4+4 would be respectively
4, 5, 0, 5, 0, 9, 0, 5, 0, 5


from above no those who is 0 or 5 can be divided by 5
hense
probability=8/10=0.8

Is this solution Helpfull? Yes (11) | No (13)  

Mohit Agarwal  (11 years ago)

any integer^4 gives 1 or 6 as unit digit except an integer containing 5 as a unit digit. according to the given condition, in all the possible cases last digit is 2,6 or 7.
Hence probability is 0.

Is this solution Helpfull? Yes (9) | No   

Sumit Rawal  (11 years ago)

The probability is 0 since X^4 will either end 1,6, or when the power is of 5 then it will end in 5 and if then x^4+1 is never divisble by by5 hence probability is 0

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rewati   (11 years ago)

multiplication by 4 gives 4,8,2,6,0 as unit digits. for a no. being divisible by 5 by adding 1 the unit digit must be either 9 or 4 . chances of meeting such condition by multiplication by 4 are 1/5 as 9 is the only no of our interest.. so.. answer is 1/5

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kapil gupta  (10 years ago)

probability will be zero as for any no x^4 last digit will not be either 4 0r 9
so 0 is answer

Is this solution Helpfull? Yes (2) | No   

Rahul Ranjan  (8 years ago)

4/5 is the only ans

Is this solution Helpfull? Yes (1) | No   

SHUBHASIS CHANDA  (11 years ago)

ans is 1/5.
the last digit can be from 1 to 9...only if the last digit is 5 or 0...the last digit will be 0 or 5....divisibility by 5 is possible...hence probability is 2/10 or 1/5

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Nimal  (11 years ago)

The unit digit of any number to the power 4 is either 1 or 6.. and even if we add 1, the resulting number will not be divisible by 5..

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Rohit Singh  (9 years ago)

for (x^4 + 1) to be divisible by 5, than it should have 0 or 5 as the last digits.
unit place of x^4 will be any of these
0^4=0;1^4=1;2^4=6;3^4=1;4^4=6;5^4=5;6^4=6;7^4=1;8^4=6;9^4=1
hence if we replace X by any of d no: from 0 to 9, non of d no: contain last digit 0 or 5.
hence probability is 0.

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