Latest Elitmus Aptitude Question SOLUTION: 'P' is a 3 digit number. First digit from the left is a square of the last digit. None of the digit is repeated in 'P'. How many number of Primesubdents are possible. (Primesubdent

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(#M40032778) ELITMUS Numerical Ability question aps Keep an EYE

'P' is a 3 digit number. First digit from the left is a square of the last digit. None of the digit is repeated in 'P'. How many number of Primesubdents are possible. (Primesubdent is a number that can not be divided by 2,3,5)
a)6
b)
c)
d)72

Asked In Elitmus Praveen kumar (10 years ago)

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Devendra Marghade  (10 years ago)

a) 6
If P=xyz, given x=z^2 Also x,y & z are different
x can be 4 & 9 i.e. square of 2 & 3 respectively,So P=4y2 & 9y3
As P shouldn't be divisible by 2,3,5 we have only P=9y3
So Combinations (913),(923),(943),(953),(973),(983) Total 6 are possible

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RAKESH  (10 years ago)

let P=xyz where x=z^2
(z,x) may be (1,1),(2,4),(3,9)
P will be of form 1y1,4y2,9y3
given that digits are not repeated so 1y1 is not possible
4y2 is also not possible bcoz its divisible by 2 so P is not Primesubdent

so P=9y3 y=0,1,2,...,9
P shouldn't be divisible by 2,3,5
so P=(913),(923),(943),(953),(973),(983)

total six numbers
a)6

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Avisek Mal  (10 years ago)

From question last digit of 'P' is 2 when first digit is 4 or last digit of 'P'

is 3 when first digit is 9.

when last digit is 2 always divided by 2.

when last digit is 3 is not divided by 3 for 913,923,943,953,973,983.

so,number of primesubdents is 6.

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saikrishna  (10 years ago)

number is not divisible by 2,3,5 so last digits in p are 1,3,7 and first digit is sqr of last it can be only 3.
the middle no can be chosen
in 10 ways the answers is 10

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