The solution is a bit flawed as it works for the limited test cases in this problem.
A classic approach was to take intersection of factors of a and b, and take their product. This works perfectly on paper. However, the inbuilt intersect() function gives unique values discarding the repeated values, thus I was not able to use it.
Swap the first and last columns
Vector of numbers divisible by 3
Maximum value in a matrix
Min of a Matrix
Last Digit of fibonacci number
Rotate a matrix without using rot90
ZigZag - 02
ZigZag - 05
Fault in our Stars - 02
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