Question

Find a 5-digit integer nnn such that n2n^2n2 starts with 27182

Answer 1

Answer:

Here is a Python (2.7) solution: print "What is a 5-digit integer N such that N^2 starts with 27182?" for n in range(10000,100000): nsquare = n*n. snsquare = str(nsquare) if (len(snsquare) > 4): s5 = "" for k in range(5): s5 = s5 + snsquare[k]

Answer 2

<h1> Hi Dude ! </ h1>

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