Question

47*xy^4 = a 5 digit perfect square and a = x-y/15 find cube root of y^a if y is a prime number

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Answer 1

Answer:

then a is not a prime numbers. and a is equal to b ok I think it should be 5

Answer 2

Given : 47xy⁴ is a five digit perfect square

a=(x-y)/15

To Find : cuberoot y^a

Solution:

47xy⁴ is a five digit perfect square

= 47x ( y²)²

Hence x  must be 47  

= 47²(y²)²

= (47y²)²

   10000   ≤ (47y²)²  < 99999

=> 100  ≤  47y² ≤ 316

=> 2  <  y² ≤  6

only  possible value of y is 2

x = 47

y = 2

a = ( x - y)/15

=> a = ( 47 - 2)/15

=> a = 3

cuberoot y^a

y = 2  , a = 3

∛2³  = 2

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