if 47×xy⁴ is a five digit perfect square and a=(x-y)/15 ,then cuberoot y^a
(y is a prime number)
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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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