Question

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If a⁵ = b⁴

c³ = d²

And (c- a ) = 19

Find the value (d - b ) = ????



Plzzz Help me Guysss.

Answer 1

a^5=b^4

This can be interpreted as

a=t^4 and b=t^5

c^3=d^2

This can be interpreted as

c=s^2 and d=s^3

Now, (c-a)=19

(s^2-t^4)=19

(s-t^2)(s+t^2)=19

Now we know that 19 is a prime number so

(s+t^2)>(s-t^2)

s+t^2=19 and s-t^2=1

=> s = 10, t = 3and so d = s^3 = 1000, b = t^5 = 243 and (d-b)=747.

Answer 2

Answer:

$\boxed{757}$

Step-by-step explanation:

Given:

a^5 = b^4

c^3 = d^2

Let p be an integer such that:

a= p^4..............(1)

Let q be an integer such that:

c = q^2.............(2)

c-a= 19

So:

q^2-p^4= 19 [From 1 and 2]

==> (q-p^2)(p^2+q)=19

The two terms (p^2+q) and (q-p^2) are integers

Multiplication of 2 integers always give an integer other than primes.

For example 2*3=6 not prime

But when a prime is multiplied by 1 then only it will give a prime number.

19 is a prime . So one of the factor is 1 and the other is 19 itself..

(p^2+q) (q-p^2)=19

Notice that p^2+q is more than p^2-q because p and q are positive...

So:

(p^2+q)(-p^2+q)=19*1

As they are both integers :

q+p^2 = 19 .............(3)

q-p^2 = 1...............(4)

Adding 3 and 4:

2 q = 20

==> q = 20/2

==> q = 10

q - p^2 = 1

-p^2 = - 9

==> p^2 = 9

==> p= 3 p cannot be negative ..

Lets recall:

c = q^2

==> c= 10^2

==> c=100

d ^2 = c^3

==> d^2 =(100)^3

==> d^2 =1000000

==> d=1000................(5)

a = p^4

= 3^4

=81

a^5 = b^4

==> 81^5 = b^4

==> b = 3^5

==> b =243

The value of d-b

=1000 -243

==> 757

The answer is 757

Hope it helps you

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