GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Moderator
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1111
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
How To Solve : Last Two Digits of Exponents Ending with 2, 4, 6, 8
[#permalink]
02 Aug 2024, 04:26
02 Aug 2024, 04:26
1
Show: ::
Attachment:
Last Two Digits of Exponents Ending with 2, 4, 6, 8
Hi All,
I have posted a video on YouTube to discuss Last Two Digits of Exponents Ending with 2, 4, 6, 8
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 \(2^{4274}\) ?
⁍ Theory of Last Two Digits of Numbers Ending with 4
⁍ Find Last two digits of \(4^{501}\) ?
⁍ Find Last two digits of \(1684^{8101}\) ?
⁍ Theory of Last Two Digits of Numbers Ending with 6
⁍ Find Last two digits of \(6^{321}\) ?
⁍ Find Last two digits of \(486^{422}\) ?
⁍ Theory of Last Two Digits of Numbers Ending with 8
⁍ Find Last two digits of \(8^{201}\) ?
⁍ Find Last two digits of \(1768^{821}\) ?
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 \(2^{4274}\) ?
Show: ::
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
Theory of Last Two Digits of Numbers Ending with 4
- • We will convert the number from \(4^{SomePower}\) to \(2^{2*SomePower}\)
• Apply the rule for power of 2 as described above
Q2. Find Last two digits of \(4^{501}\) ?
Show: ::
Sol: \(4^{501}\) = \(2^{2*501}\)
= \(2^{1002}\)
= \(2^{1000 + 2}\)
= \(2^{10*100} * 2^2\)
= \((2^{10})^{100} * 4\)
= \((1024)^{Even} * 4\)
= \(76 * 4\)
= 04
=> Last two digits = 04
= \(2^{1002}\)
= \(2^{1000 + 2}\)
= \(2^{10*100} * 2^2\)
= \((2^{10})^{100} * 4\)
= \((1024)^{Even} * 4\)
= \(76 * 4\)
= 04
=> Last two digits = 04
Q3. Find Last two digits of \(1684^{8101}\) ?
Show: ::
Sol: \((421*4)^{8101}\)
= \((421)^{8101} * 2^{2*8101}\)
= 21 * \(2^{16202}\)
= 21 * \(2^{16200 + 2}\)
= 21 * \(2^{10*1620}* 2^2\)
= 21 * \((2^{10})^{1620}* 4\)
= 21 * \((1024)^{Even}* 4\)
= 84 * 76
= 84
=> Last two digits = 84
= \((421)^{8101} * 2^{2*8101}\)
= 21 * \(2^{16202}\)
= 21 * \(2^{16200 + 2}\)
= 21 * \(2^{10*1620}* 2^2\)
= 21 * \((2^{10})^{1620}* 4\)
= 21 * \((1024)^{Even}* 4\)
= 84 * 76
= 84
=> Last two digits = 84
Theory of Last Two Digits of Numbers Ending with 6
- • We will convert the number from \(6^{SomePower}\) to \(2^{SomePower} * 3^{SomePower}\)
• Apply the rule for power of 2 and 3 to get the answer
Q4. Find Last two digits of \(6^{321}\) ?
Show: ::
Sol: \(6^{321}\) = \(2^{321} * 3^{321}\)
= \(2^{320 + 1} * 3^{320 + 1}\)
= \(2^{10*32} * 2^1 * 3^{4*80} * 3^1\)
= \((2^{10})^{32} * 2 * (3^{4})^{80} * 3\)
= \((1024)^{Even} * 2 * (81)^{80} * 3\)
= \(76 * 6 * 01\)
= 56
=> Last two digits = 56
= \(2^{320 + 1} * 3^{320 + 1}\)
= \(2^{10*32} * 2^1 * 3^{4*80} * 3^1\)
= \((2^{10})^{32} * 2 * (3^{4})^{80} * 3\)
= \((1024)^{Even} * 2 * (81)^{80} * 3\)
= \(76 * 6 * 01\)
= 56
=> Last two digits = 56
Q5. Find Last two digits of \(486^{422}\) ?
Show: ::
Sol: \((243*2)^{422}\)
= \((3^5)^{422} * 2^{420 + 2}\)
= \(3^{5 *422} * 2^{10*42} * 2^2\)
= \(3^{2110} * 1024^{Even} * 4\)
= \(3^{2108+2} * 76 * 4\)
= \(3^{4 * 527} * 3^2 * 04\)
= \(81^{527} * 9 * 04\)
= \(61 * 36\)
= 96
=> Last two digits = 96
= \((3^5)^{422} * 2^{420 + 2}\)
= \(3^{5 *422} * 2^{10*42} * 2^2\)
= \(3^{2110} * 1024^{Even} * 4\)
= \(3^{2108+2} * 76 * 4\)
= \(3^{4 * 527} * 3^2 * 04\)
= \(81^{527} * 9 * 04\)
= \(61 * 36\)
= 96
=> Last two digits = 96
Theory of Last Two Digits of Numbers Ending with 8
- • We will convert the number from \(8^{SomePower}\) to \(2^{3*SomePower}\)
• Apply the rule for power of 2 as described above
Q6. Find Last two digits of \(8^{201}\) ?
Show: ::
Sol: \(8^{201}\) = \(2^{3*201}\)
= \(2^{603}\)
= \(2^{600 + 3}\)
= \(2^{10*60} * 2^3\)
= \((2^{10})^{60} * 8\)
= \((1024)^{Even} * 8\)
= \(76 * 8\)
= 08
=> Last two digits = 08
= \(2^{603}\)
= \(2^{600 + 3}\)
= \(2^{10*60} * 2^3\)
= \((2^{10})^{60} * 8\)
= \((1024)^{Even} * 8\)
= \(76 * 8\)
= 08
=> Last two digits = 08
Q7. Find Last two digits of \(1768^{821}\) ?
Show: ::
Sol: \((221*8)^{821}\)
= \((221)^{821} * 2^{3*821}\)
= 21 * \(2^{2463}\)
= 21 * \(2^{2460 + 3}\)
= 21 * \(2^{10*246}* 2^3\)
= 21 * \((2^{10})^{246}* 8\)
= 21 * \((1024)^{Even}* 8\)
= 68 * 76
= 68
=> Last two digits = 68
= \((221)^{821} * 2^{3*821}\)
= 21 * \(2^{2463}\)
= 21 * \(2^{2460 + 3}\)
= 21 * \(2^{10*246}* 2^3\)
= 21 * \((2^{10})^{246}* 8\)
= 21 * \((1024)^{Even}* 8\)
= 68 * 76
= 68
=> Last two digits = 68
Link to Theory for Last Two digits of exponents here.
Link to Theory for Units' digit of exponents here.
Hope it helps!
gmatclubot
How To Solve : Last Two Digits of Exponents Ending with 2, 4, 6, 8 [#permalink]
02 Aug 2024, 04:26
Moderators:
|
|
||
