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anand bhatnagar

find last two digit

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20.  The last two digits each nn number follows a specific pattern.
There is likely a more simple way of addressing this pattern...
If the final (ones) digit of n were 0, 1 or 9, then the final digit of nn is respectively 0, 1 or 9. If the final digit of n were 2 or 8, then the final digit of nn is respectively 4 and 6 where the tens-digit is even,or 6 and 4 if.odd. If the final digit of n were 3 or 7, then the final digit of nn is respectively 7 and 3 where the tens-digit is even,or 3 and 7 if.odd. In each of the above cases, the tens-digit is represented respectively by each different numeral, i.e, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} for each n. The sum of these ten digits will be 45, and with the tens digit of the sum being 0 (5 × 6 = 30) as the left-most digit of the six pairs is a hundreds digit. If the final digit of n were 4 or 6, then the final digit of nn is respectively  6. The tens-digit, in both follows the sequence {5, 1, 7, 3, 9}. The sum of the four sets of the quintuplets of tens-digits is (4 × (5+1+7+3+9) = 200). And the sum of the twenty ones-digits ending in 6 is 120. The total sum for the final digits of 4 and 6, then is 320; that is, the ones-digit is 0 and the tens-digit is 2. If the final digit of n were 5, then the final digit of nn is respectively 5. The tens-digit of nn alternates between 2 for the even tens-digit of n and 7 for the odd tens-digit of n. Given the five pairs, the sum of these ten is 500, that is, the ones-digit and tens-digit are both 0. Thus, by summation, the ones-digit in the final solution must be 0. and the tens-digit must be a 2.    


 

Edited by DejMar
correction of typo

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Spoiler

20

 

because python

 

def nnmod100(n):
    nm = n % 100;
    p = 1;
    for i in xrange(n):
        p = (p*nm) % 100;
    return p;
s = sum(nnmod100(n) for n in range(1,100));
print s % 100;
   
 

 

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