Question
Determine the number of trailing zeros in the product of
the following numbers: 11 × 22 × 33 × … × 2222Solution
Total number of 5’s in the given expression: 55 = 15 × 55 = 5 1010 = 210 × 510 = 10 1515 = 315 × 515 = 15 2020 = 420 × 520 = 20 Total number of 2’s in the given expression: 22 = 2 44 = (22)4 = 28 = 8 66 = (2 × 3)6 = 26 × 36 = 6 88 = (23)8 = 224 = 24 1010 = 210 × 510 = 10 So, total number of 5’s in the given expression = 5 + 10 + 15 + 20 = 50 And, at least 50 2’s in the given expression So, number of zeroes at the end of the product = 50Â
IRDA has the power to frame the regulations under section ___of the Insurance Act. 1938.
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