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Number less than equals to N with maximum product of prime factors | Function to find the smallest number having a maximum product of prime factors.
Declare the list arr.
Initialise list with 1.
Iterate from [ 2 , N ].
If value at index i is 1 , then i is prime and make update list at index for all multiples of i.
Initialise maxValue.
Find number having maximum product of prime factor <= N.
Find the maximum value.
Given number.
Function call"
Solution in Python:
 What is a code to number less than equals to N with maximum product of prime factors and the other steps required?
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#Here is code to number less than equals to N with maximum product of prime factors and the other steps required.
 def maxPrimefactorNum ( N ) :
	arr = [ ]
	for i in range ( N + 1 ) :
		arr . append ( 1 )
	for i in range ( 2 , N + 1 ) :
		if ( arr [ i ] == 1 ) :
			for j in range ( i , N + 1 , i ) :
				arr [ j ] *= i
	maxValue = 1
	for i in range ( 2 , N + 1 ) :
		if ( arr [ i ] > maxValue ) :
			maxValue = i
	return maxValue
N = 20 ;
print ( maxPrimefactorNum ( N ) )