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Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1115
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
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WE:Engineering (Computer Software)
How to Solve: Last Two Digits of Exponents Ending with 1, 3, 7, 9
[#permalink]
26 Jul 2024, 08:09
26 Jul 2024, 08:09
1
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Attachment:
Last Two Digits of Exponents Ending with 1, 3, 7, 9
Hi All,
I have posted a video on YouTube to discuss Last Two Digits of Exponents Ending with 1, 3, 7, 9
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 1
⁍ Find Last two digits of \(131^{345}\) ?
⁍ Theory of Last Two Digits of Numbers Ending with 3
⁍ Find Last two digits of \(3^{241}\) ?
⁍ Find Last two digits of \(783^{402}\) ?
⁍ Theory of Last Two Digits of Numbers Ending with 7
⁍ Find Last two digits of \(7^{282}\) ?
⁍ Find Last two digits of \(847^{422}\) ?
⁍ Theory of Last Two Digits of Numbers Ending with 9
⁍ Find Last two digits of \(9^{243}\) ?
⁍ Find Last two digits of \(1269^{436}\) ?
Theory of Last Two Digits of Numbers Ending with 1
- • Units' digit of the number = 1
• Tens' digit of the number = Tens' digit of the base * Units' digit of the exponent
Q1. Find Last two digits of \(131^{345}\) ?
Show: ::
Sol: Base = 131
Exponent = 345
=> Units' digit = 1
=> Tens' digit = 3 * 5 [131 * 345]
= 5
=> Last two digits = 51
Exponent = 345
=> Units' digit = 1
=> Tens' digit = 3 * 5 [131 * 345]
= 5
=> Last two digits = 51
Theory of Last Two Digits of Numbers Ending with 3
- • \(3^4 = 81\)
• We need to express the power of two into product of \(3^{MultipleOf 4 Power}\) * \(3^{SmallerPower}\)
• We will have last two digits as \(81^{SomePower}\) * \(3^{SmallerPower}\)
• We can use Logic of Last Two Digits of Exponents ending with 1 * last two digits of \(3^{SmallerPower}\)
Q2. Find Last two digits of \(3^{241}\) ?
Show: ::
Sol: \(3^{241}\) = \(3^{240 + 1}\)
= \(3^{4*60} * 3^1\)
= \((3^4)^{60} * 3\)
= \((81)^{60} * 3\)
= 01 * 3 = 03
=> Last two digits = 03
= \(3^{4*60} * 3^1\)
= \((3^4)^{60} * 3\)
= \((81)^{60} * 3\)
= 01 * 3 = 03
=> Last two digits = 03
Q3. Find Last two digits of \(783^{402}\) ?
Show: ::
Sol: \((261*3)^{402}\)
= \((261)^{402} * 3^{400 + 2}\)
= 21 * \((3^4)^{100} * 3^2\)
= 21 * \((81)^{100} * 9\)
= 21 * 01 * 9
= 89
=> Last two digits = 89
= \((261)^{402} * 3^{400 + 2}\)
= 21 * \((3^4)^{100} * 3^2\)
= 21 * \((81)^{100} * 9\)
= 21 * 01 * 9
= 89
=> Last two digits = 89
Theory of Last Two Digits of Numbers Ending with 7
- • Last two digits of \(7^4 = 01\)
• We need to express the power of two into product of \(7^{MultipleOf 4 Power}\) * \(7^{SmallerPower}\)
• We will have last two digits as \(01^{SomePower}\) * \(7^{SmallerPower}\)
• We can use Logic of Last Two Digits of Exponents ending with 1 * last two digits of \(7^{SmallerPower}\)
Q4. Find Last two digits of \(7^{282}\) ?
Show: ::
Sol: \(7^{282}\) = \(7^{280 + 2}\)
= \(7^{4*70} * 7^2\)
= \((7^4)^{70} * 49\)
= \((01)^{70} * 49\)
= 01 * 49 = 49
=> Last two digits = 49
= \(7^{4*70} * 7^2\)
= \((7^4)^{70} * 49\)
= \((01)^{70} * 49\)
= 01 * 49 = 49
=> Last two digits = 49
Q5. Find Last two digits of \(847^{422}\) ?
Show: ::
Sol: \((121*7)^{422}\)
= \((121)^{422} * 7^{420 + 2}\)
= 41 * \((7^4)^{107} * 7^2\)
= 41 * \((01)^{107} * 49\)
= 41 * 01 * 49
= 09
=> Last two digits = 09
= \((121)^{422} * 7^{420 + 2}\)
= 41 * \((7^4)^{107} * 7^2\)
= 41 * \((01)^{107} * 49\)
= 41 * 01 * 49
= 09
=> Last two digits = 09
Theory of Last Two Digits of Numbers Ending with 9
- • Last two digits of \(9^2 = 81\)
• We need to express the power of two into product of \(9^{Even Power}\) * \(9^{SmallerPower}\)
• We will have last two digits as \(81^{SomePower}\) * \(9^{SmallerPower}\)
• We can use Logic of Last Two Digits of Exponents ending with 1 * last two digits of \(9^{SmallerPower}\)
Q6. Find Last two digits of \(9^{243}\) ?
Show: ::
Sol: \(9^{242 + 1}\)
= \(9^{2*121} * 9^1\)
= \((9^2)^{121} * 9\)
= \((81)^{121} * 9\)
= 81 * 9 = 29
=> Last two digits = 29
= \(9^{2*121} * 9^1\)
= \((9^2)^{121} * 9\)
= \((81)^{121} * 9\)
= 81 * 9 = 29
=> Last two digits = 29
Q7. Find Last two digits of \(1269^{436}\) ?
Show: ::
Sol: \((141*9)^{436}\)
= \((141)^{436} * 9^{2*218}\)
= 41 * \((9^2)^{218}\)
= 41 * \((81)^{218}\)
= 41 * 41
= 81
=> Last two digits = 81
= \((141)^{436} * 9^{2*218}\)
= 41 * \((9^2)^{218}\)
= 41 * \((81)^{218}\)
= 41 * 41
= 81
=> Last two digits = 81
Link to Theory for Last Two digits of exponents here.
Link to Theory for Units' digit of exponents here.
Hope it helps!
Re: How to Solve: Last Two Digits of Exponents Ending with 1, 3, 7, 9
[#permalink]
26 Jul 2024, 10:20
26 Jul 2024, 10:20
Expert Reply
gmatclubot
Re: How to Solve: Last Two Digits of Exponents Ending with 1, 3, 7, 9 [#permalink]
26 Jul 2024, 10:20
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