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I have posted a video on YouTube to discuss Last Two Digits of Exponents Ending with 2
Attached pdf of this Article as SPOILER at the top! Happy learning!
Following is Covered in the Video
Theory of Last Two Digits of Numbers Ending with 2
⁍ Find Last two digits of \(2^{4274}\) ? ⁍ Find Last two digits of \(842^{9802}\) ?
Theory of Last Two Digits of Numbers Ending with 2
• Express the Number as \((2^{10})^{Power}\) * \(2^{Smaller Power}\) • Now we know that \(2^{10}\) = 1024 and we have expressed the number \(1024^{Power}\) • \(24^{Odd Power}\) will have last two digits as 24 • \(24^{Even Power}\) will have last two digits as 76 • If we have power of power then we can use last two digits of \(76^{Any Positive Integer}\) is 76
Sol: \(842^{9802}\) = \((421 * 2)^{9802}\) = \((421)^{9802} * 2^{9800 + 2}\) => Last two digits = 41 * Last two digits of \((2^{10})^{980} * 2^2\) [Watch this video to learn about How to Find Last two digits of Exponents ending with 1] => Last two digits = 41 * Last two digits of \(1024^{Even} * 4\) => Last two digits = 41 * Last two digits of 76 * 4 => Last two digits = 64