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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
[tdo=2]We need to find the cycle of last two digits of 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
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
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
Q3. Find Last two digits of \(7^{80a + 51 }\) (given that a is a positive integer)?
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