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Joined: 02 Jan 2020
Status:GRE Quant Tutor
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GMAT 1: 700 Q51 V31
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How to Solve: Last Two Digits of Power of 6
[#permalink]
21 Sep 2024, 00:44
21 Sep 2024, 00:44
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How to Solve: Last Two Digits of Power of 6
Hi All,
I have posted a video on YouTube to discuss Last Two Digits of Power of 6
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 6
• Find Units’ digit of \(6^{71}\) ?
• Find Units’ digit of \(6^{12517}\) ?
Theory of Last Two Digits of Power of 6
• To find Last Two Digits of any positive integer power of 6
| We need to find the cycle of last two digits of power of 6 | |
| \(6^1\) last two digits is 06 \(6^2\) last two digits is 06*6 = 36 \(6^3\) last two digits is 36*6 = 16 \(6^4\) last two digits is 16*6 = 96 \(6^5\) last two digits is 96*6 = 76 \(6^6\) last two digits is 76*6 = 56 | \(6^7\) last two digits is 56*6 = 36 \(6^8\) last two digits is 36*6 = 16 |
=> The power repeats after every \(5^{th}\) power
=> Cycle of last two digits of power of 6 = 5 (ignoring \(6^1\))
=> We need to subtract one from the power and then divide the power by 5 and check the remainder
=> Last two digits will be same as last two digits of \(6^{1 + Remainder}\)
NOTE: If Remainder is 0 then last two digits = last two digits of \(6^{1 + Cycle}\) = last two digits of \(6^{6}\) = 56
Q1. Find Last two digits of \(6^{71}\)?
Show: ::
Sol: We need to subtract 1 from 71 and then divided the remaining power (71-1=70) by 5 and get the remainder
70 divided by 5 gives 0 remainder = remainder of 5
=> Last two digits of \(6^{71}\) = Last two digits of \(6^{1+5}\) = 56
70 divided by 5 gives 0 remainder = remainder of 5
=> Last two digits of \(6^{71}\) = Last two digits of \(6^{1+5}\) = 56
Q2. Find Last two digits of \(6^{12517}\)?
Show: ::
Sol: 12517 - 1 = 12516
12516 divided by 5 will give the same remainder as 6 by 5 which is 1
Watch this video to Master Divisibility Rules
=> Last two digits of \(6^{12517}\) = Last two digits of \(6^{1+1}\) = 36
12516 divided by 5 will give the same remainder as 6 by 5 which is 1
Watch this video to Master Divisibility Rules
=> Last two digits of \(6^{12517}\) = Last two digits of \(6^{1+1}\) = 36
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 6 [#permalink]
21 Sep 2024, 00:44
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