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Joined: 02 Jan 2020
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How to Solve: Last Two Digits of Power of 11
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27 Sep 2024, 05:01
27 Sep 2024, 05:01
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How to Solve: Last Two Digits of Power of 11
Hi All,
I have posted a video on YouTube to discuss Last Two Digits of Power of 11
Attached pdf of this Article as SPOILER at the top! Happy learning!
Following is Covered in the Video
- • Theory of Last Two Digits of Power of 11
• Find Units’ digit of \(11^{128}\) ?
• Find Units’ digit of \(11^{15342}\) ?
Theory of Last Two Digits of Power of 11
• To find Last Two Digits of any positive integer power of 11
| We need to find the cycle of last two digits of power of 11 | |
| \(11^1\) last two digits is 11 \(11^2\) last two digits is 11*11 = 21 \(11^3\) last two digits is 21*11 = 31 \(11^4\) last two digits is 31*11 = 41 \(11^5\) last two digits is 41*11 = 51 \(11^6\) last two digits is 51*11 = 61 | \(11^7\) last two digits is 61*11 = 71 \(11^8\) last two digits is 71*11 = 81 \(11^9\) last two digits is 81*11 = 91 \(11^10\) last two digits is 91*11 = 01 \(11^11\) last two digits is 01*11 = 11 \(11^12\) last two digits is 11*11 = 21 |
=> The power repeats after every \(10^{th}\) power
=> Cycle of last two digits of power of 11 = 10
=> We need to divide the power by 10 and check the remainder
=> Last two digits will be same as last two digits of \(11^{Remainder}\)
=> Unit's digit is 1, Tens' digit = Remainder
NOTE: If Remainder is 0 then last two digits = last two digits of \(11^{Cycle}\) = last two digits of \(11^{10}\) = 01
Q1. Find Last two digits of \(11^{128}\)?
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Sol: We need to divided the power (128) by 10 and get the remainder
128 divided by 10 gives 8 remainder
Watch this video to Master Divisibility Rules
=> Last two digits of \(11^{128}\) = Last two digits of \(11^8\) = 81
128 divided by 10 gives 8 remainder
Watch this video to Master Divisibility Rules
=> Last two digits of \(11^{128}\) = Last two digits of \(11^8\) = 81
Q2. Find Last two digits of \(11^{15342}\)?
Show: ::
Sol: 15342 divided by 10 will give the same remainder as 2 by 10 which is 2
Watch this video to Master Divisibility Rules
=> Last two digits of \(11^{15342}\) = Last two digits of \(11^2\) = 21
Watch this video to Master Divisibility Rules
=> Last two digits of \(11^{15342}\) = Last two digits of \(11^2\) = 21
Hope it helps!
Link to Theory for Last Two digits of exponents here.
Link to Theory for Units' digit of exponents here.
gmatclubot
How to Solve: Last Two Digits of Power of 11 [#permalink]
27 Sep 2024, 05:01
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