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Joined: 02 Jan 2020
Status:GRE Quant Tutor
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How to Solve: Last Two Digits of Power of 7
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14 Sep 2024, 06:15
14 Sep 2024, 06:15
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How to Solve: Last Two Digits of Power of 7
Hi All,
I have posted a video on YouTube to discuss Last Two Digits of Power of 7
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 7
• Find Units’ digit of \(7^{93}\) ?
• Find Units’ digit of \(7^{1529}\) ?
• Find Units’ digit of \(7^{80a + 51}\) (given that a is a positive integer)?
Theory of Last Two Digits of Power of 7
• To find Last Two Digits of any positive integer power of 7
| \(7^1\) last two digits is 07 \(7^2\) last two digits is 07*7 = 49 \(7^3\) last two digits is 49*7 = 43 \(7^4\) last two digits is 43*7 = 01[/b] | \(7^5\) last two digits is 01*7 = 07 \(7^6\) last two digits is 07*7 = 49 \(7^7\) last two digits is 49*7 = 43 \(7^8\) last two digits is 43*7 = 01 |
=> The power repeats after every \(4^{th}\) power
=> Cycle of last two digits of power of 7 = 4
=> We need to divide the power by 4 and check the remainder
=> Last two digits will be same as last two digits of \(7^{Remainder}\)
NOTE: If Remainder is 0 then last two digits = last two digits of \(7^{Cycle}\) = last two digits of \(7^{4}\) = 01
Q1. Find Last two digits of \(7^{93}\)?
Show: ::
Sol: We need to divided the power (93) by 4 and get the remainder
93 divided by 4 gives 1 remainder
=> Last two digits of \(7^{93}\) = Last two digits of \(7^1\) = 07
93 divided by 4 gives 1 remainder
=> Last two digits of \(7^{93}\) = Last two digits of \(7^1\) = 07
Q2. Find Last two digits of \(7^{1529}\)?
Show: ::
Sol: 1529 divided by 4 will give the same remainder as 29 by 4 which is 1
Watch this video to Master Divisibility Rules
=> Last two digits of \(7^{1529}\) = Last two digits of \(7^1\) = 07
Watch this video to Master Divisibility Rules
=> Last two digits of \(7^{1529}\) = Last two digits of \(7^1\) = 07
Q3. Find Last two digits of \(7^{80a + 51 }\) (given that a is a positive integer)?
Show: ::
Sol: Remainder of 80a + 51 divided by 4 = Remainder of 80a by 4 + Remainder of 51 by 4
= 0 + 3 = 3
=> Last two digits of \(7^{80a + 51}\) = Last two digits of \(7^3\) = 43
= 0 + 3 = 3
=> Last two digits of \(7^{80a + 51}\) = Last two digits of \(7^3\) = 43
Hope it helps!
Link to Theory for Last Two digits of exponents here.
Link to Theory for Units' digit of exponents here.
Re: How to Solve: Last Two Digits of Power of 7
[#permalink]
14 Sep 2024, 23:06
14 Sep 2024, 23:06
Expert Reply
Added to the long lasting, at the moment
GRE math handout project
https://gre.myprepclub.com/forum/gre-ma ... tml#p95352
GRE math handout project
https://gre.myprepclub.com/forum/gre-ma ... tml#p95352
gmatclubot
Re: How to Solve: Last Two Digits of Power of 7 [#permalink]
14 Sep 2024, 23:06
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