The last two digit of the multiplication of 1357.........9799 will be

Question

Answer 1

1*3*5*...*99 = 25k
25k mod 100 = 25 if k mod 4 = 1
25k mod 100 = 75 if k mod 4 = 3
In this case, 1*3*...*23*27*...*99 mod 4 = 1*-1*1*-1*...*1 (25 pairs) = -1 = 3
So, remainder = 75

Answer 2

1*3*5*...*99 = 25k
25k mod 100 = 25 if k mod 4 = 1
25k mod 100 = 75 if k mod 4 = 3
In this case, 1*3*...*23*27*...*99 mod 4 = 1*-1*1*-1*...*1 (25 pairs) = -1 = 3
So, remainder = 75

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The last two digit of the multiplication of 1*3*5*7.........97*99 will be

The last two digit of the multiplication of 1357.........9799 will be