Latest TCS Aptitude Question SOLUTION: Given that a,b,c,d are positive integers such that b^3/2=a and d^5/4 =c. If b-d= 9 then find the value of a-c

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Given that a,b,c,d are positive integers such that b^3/2=a and d^5/4 =c. If b-d= 9 then find the value of a-c

Asked In TCS (11 years ago)

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Piyush Jain  (11 years ago)

From hit n trial method
we get
b=25 and d=16
((5)^(2*3/2))=a
a=125
((2)^(4*5/4))=c
c=32
answer 125-32=93

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lokesh daiya  (11 years ago)

b=a^2/3
d=c^4/5
now b-d=9
a^2/3-c^4/5=9
(a^1/3+c^2/3)(a^1/3-c^2/3)=3*3 ---------by (x^2-y^2=(x+y)(x-y))
so a^1/3+c^2/3=3
and a^1/3-c^2/3=3
add both expression
then 2a^1/3=6
a^1/3=3
a=27
putting this value in upper expression we get c=0
so answer is a-c=27
here u can see that value of a=27,b=9,c=0,d=0

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sainath  (11 years ago)

NO such value is possible with satisfying a,b,c,d to be INTEGERS.

Because, given b-d = 9 which is an odd Integer.
So one of b,d must be odd and one even Integer. [like (11,2),(13,4).,]
[since difference between 2 odd or 2 even integers is an even Integer]
Multiplication of two odd integers results an odd integer.so if 'a' is an odd
Integer then a^b is an odd Integer.
So one of b^3 , d^5 is an odd integer.

But according to a,c must be integers,
b^3 must be divisible by '2'and d^5 must be divisible by '4'.
One of them is impossible BECAUSE 2 or multiple of 2 cant divide an odd number perfectly.

SO THERE IS NO ANSWER FOR THIS QUESTION.

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