Find the smallest value of k, when, a) 280k is a perfect square, b) 882k is a cube.

Here we have to find the smallest value of k for which
280k is a perfect square and 882k is a perfect cube.

Now,
writing 280 as a product of prime numbers, we get 7*2*2*2*5. Writing 882 as a product of
prime numbers we get 3*3*2*7*7.

Now, for 280k to be a
perfect square all the prime factors should be in pairs, and if 882k has to be a perfect
cube all the prime factors should be in triplets. Or in other words 280k should have a
square of every relevant prime number and 882k should have a cube of the same prime
numbers.

For this, we need k to have one factor equal to 32
as 32*8 is a square and 32*2 is a cube. Also k needs to have another factor of 7 as 7*7
is a square and 7*49 is a cube. Similarly, k needs to have a factor of 125 as 125*5 is a
square and 125*1 is a cube. And finally k needs to have a factor of 81 as 81*1 is a
square and 81*9 is a cube.

In this way we
arrive at the factors of k as 32, 81, 125 and 7. So k is equal to
2268000.

We see that 280*2268000 is a
perfect square and 882*2268000 is a perfect cube. This is not possible with any value of
k smaller than 2268000.