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Joined: 02 Jan 2020
Status:GRE Quant Tutor
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Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
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How to Solve: Last Two Digits of Exponents Ending with 2
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22 Jul 2024, 04:34
22 Jul 2024, 04:34
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Last Two Digits of Exponents Ending with 2
Hi All,
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
Q1. Find Last two digits of \(2^{4274}\) ?
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Sol: \(2^{4274}\) = \(2^{4270 + 4}\) = \(2^{10 * 427} * 2^4\)
= \((2^{10})^{ 427} * 16\) = \(1024^{427} * 16\)
= \(1024^{Odd} * 16\)
=> Last two digits 24 * 16
=> Last two digits = 84
= \((2^{10})^{ 427} * 16\) = \(1024^{427} * 16\)
= \(1024^{Odd} * 16\)
=> Last two digits 24 * 16
=> Last two digits = 84
Q2. Find Last two digits of \(842^{9802}\) ?
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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
= \((421)^{9802} * 2^{9800 + 2}\)
=> 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
Link to Theory for Last Two digits of exponents here.
Link to Theory for Units' digit of exponents here.
Hope it helps!
Re: How to Solve: Last Two Digits of Exponents Ending with 2
[#permalink]
22 Jul 2024, 04:50
22 Jul 2024, 04:50
Expert Reply
Hello
gmatclubot
Re: How to Solve: Last Two Digits of Exponents Ending with 2 [#permalink]
22 Jul 2024, 04:50
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