The number of zeroes at the end of is

JEE MainMathematicsToughBinomial Theorem

The number of zeroes at the end of $(101)^{11} - 1$ is

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(101)^{11} - 1 = (1 + 100)^{11} - 1

[IMAGE] = 100 (odd number)
The number of zeros of the end is

Source: Grass-G / M-20 Mathematics Paper 12 Dec.docx

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