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Attachment:
Units Digit of Power of 4.pdf [158.34 KiB]
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How to Solve: Units' Digit of Power of 4


Hi All,

I have posted a video on YouTube to discuss Units' Digit of Power of 4



Attached pdf of this Article as SPOILER at the top! Happy learning! :)

Following is Covered in the Video

Theory of Units' Digit of Power of 4
    ⁍ Find Units’ digit of \(4^{51}\) ?
    ⁍ Find Units’ digit of \(4^{33}\) ?
    ⁍ Find Units’ digit of \(4^{44}\) ?
    ⁍ Find Units’ digit of \(4^{60x + 61}\) (given that x is a positive integer)?
    ⁍ Find Units’ digit of \(12954^{1053}\) ?


Theory of Units' Digit of Power of 4

• To find units' digit of any positive integer power of 4


We need to find the cycle of units' digit of power of 4
\(4^1\) units’ digit is 4
\(4^2\) units’ digit is 6
\(4^3\) units’ digit is 4
\(4^4\) units’ digit is 6

=> The power repeats after every \(2^{nd}\) power
=> Cycle of units' digit of power of 4 = 2

=> Units' digit of odd power of 4 = 4
=> Units' digit of even power of 4 = 6

Q1. Find Units’ digit of \(4^{51}\)?

Sol: 51 is Odd
=> Units' digit of \(4^{51}\) = 4

Q2. Find Units’ digit of \(4^{33}\)?

Sol: 33 is Odd
=> Units' digit of \(4^{33}\) = 4

Q3. Find Units’ digit of \(4^{44}\)?

Sol: 44 is Even
=> Units' digit of \(4^{Even}\) = 6

Q4. Find Units’ digit of \(4^{60x + 61}\) (given that x is a positive integer)?

Sol: 60x + 61 = Even + Odd = Odd
=> Units' digit of \(4^{60x + 61}\) = 4

Q5. Find Units’ digit of \(12954^{1053}\) ?

Sol: Units' digit of power of any number = Units' digit of power of the units' digit of that number
=> Units’ digit of \(12954^{1053}\) = Units’ digit of \(4^{1053}\)
=> 1053 is Odd
=> Units' digit of \(12954^{1053}\) = Units, digit of \(4^{1053}\)= 4

Hope it helps!
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Attachment:
Units Digit of Power of 5.pdf [162.84 KiB]
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How to Solve: Units' Digit of Power of 5


Hi All,

I have posted a video on YouTube to discuss Units' Digit of Power of 5



Attached pdf of this Article as SPOILER at the top! Happy learning! :)

Following is Covered in the Video

Theory of Units' Digit of Power of 5
    ⁍ Find Units’ digit of \(5^{61}\) ?
    ⁍ Find Units’ digit of \(5^{33}\) ?
    ⁍ Find Units’ digit of \(5^{79x + 31}\) (given that x is a positive integer)?
    ⁍ Find Units’ digit of \(1055^{199}\) ?


Theory of Units' Digit of Power of 5

• To find units' digit of any positive integer power of 5


We need to find the cycle of units' digit of power of 5
\(5^1\) units’ digit is 5\(5^2\) units’ digit is 5

=> Units’ digit of any positive integer power of 5 = 5

Q1. Find Units’ digit of \(5^{61}\)?

Sol: Since 61 is a positive integer
=> Units' digit of \(5^{61}\) = 5

Q2. Find Units’ digit of \(5^{33}\)?

Sol: Since 33 is a positive integer
=> Units' digit of \(5^{33}\) = 5

Q3. Find Units’ digit of \(5^{79x + 31}\) (given that x is a positive integer)?

Sol: Since 79x + 31 is a positive integer
=> Units' digit of \(5^{79x + 31}\) = 5

Q4. Find Units’ digit of \(1055^{199}\) ?

Sol: Units' digit of power of any number = Units' digit of power of the units' digit of that number
=> Units’ digit of \(1055^{199}\) = Units’ digit of \(5^{199}\)
Since 199 is a positive integer
=> Units' digit of \(1055^{199}\) = 5

Hope it helps!
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Units Digit of Power of 6.pdf [162.88 KiB]
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How to Solve: Units' Digit of Power of 6


Hi All,

I have posted a video on YouTube to discuss Units' Digit of Power of 6



Attached pdf of this Article as SPOILER at the top! Happy learning! :)

Following is Covered in the Video

Theory of Units' Digit of Power of 6
    ⁍ Find Units’ digit of \(6^{71}\) ?
    ⁍ Find Units’ digit of \(6^{54}\) ?
    ⁍ Find Units’ digit of \(6^{70x + 41}\) (given that x is a positive integer)?
    ⁍ Find Units’ digit of \(2756^{205}\) ?


Theory of Units' Digit of Power of 6

• To find units' digit of any positive integer power of 6


We need to find the cycle of units' digit of power of 6
\(6^1\) units’ digit is 6\(6^2\) units’ digit is 6

=> Units’ digit of any positive integer power of 6 = 6

Q1. Find Units’ digit of \(6^{71}\)?

Sol: Since 71 is a positive integer
=> Units' digit of \(6^{71}\) = 6

Q2. Find Units’ digit of \(6^{54}\)?

Sol: Since 54 is a positive integer
=> Units' digit of \(6^{54}\) = 6

Q3. Find Units’ digit of \(6^{70x + 41}\) (given that x is a positive integer)?

Sol: Since 70x + 41 is a positive integer
=> Units' digit of \(6^{70x + 41}\) = 6

Q4. Find Units’ digit of \(2756^{205}\) ?

Sol: Units' digit of power of any number = Units' digit of power of the units' digit of that number
=> Units’ digit of \(2756^{205}\) = Units’ digit of \(6^{205}\)
Since 205 is a positive integer
=> Units' digit of \(2756^{205}\) = 6

Hope it helps!
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Attachment:
Units Digit of Power of 7.pdf [185.38 KiB]
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How to Solve: Units' Digit of Power of 7


Hi All,

I have posted a video on YouTube to discuss Units' Digit of Power of 7



Attached pdf of this Article as SPOILER at the top! Happy learning! :)

Following is Covered in the Video

Theory of Units' Digit of Power of 7
    ⁍ Find Units’ digit of \(7^{81}\) ?
    ⁍ Find Units’ digit of \(7^{37}\) ?
    ⁍ Find Units’ digit of \(7^{52}\) ?
    ⁍ Find Units’ digit of \(7^{80a + 51}\) (given that a is a positive integer)?
    ⁍ Find Units’ digit of \(1297^{2041}\) ?


Theory of Units' Digit of Power of 7

• To find units' digit of any positive integer power of 7


We need to find the cycle of units' digit of power of 7
\(7^1\) units’ digit is 7
\(7^2\) units’ digit is 9
\(7^3\) units’ digit is 3
\(7^4\) units’ digit is 1
\(7^5\) units’ digit is 7
\(7^6\) units’ digit is 9
\(7^7\) units’ digit is 3
\(7^8\) units’ digit is 1

=> The power repeats after every \(4^{th}\) power
=> Cycle of units' digit of power of 7 = 4
=> We need to divide the power by 4 and check the remainder
=> Units' digit will be same as Units' digit of \(7^{Remainder}\)

NOTE: If Remainder is 0 then units' digit = units' digit of \(7^{Cycle}\) = units' digit of \(7^{4}\) = 1

Q1. Find Units’ digit of \(7^{81}\)?

Sol: We need to divided the power (81) by 4 and get the remainder
81 divided by 4 gives 1 remainder
=> Units' digit of \(7^{81}\) = Units' digit of \(7^1\) = 7

Q2. Find Units’ digit of \(7^{37}\)?

Sol: 37 divided by 4 gives 1 remainder
=> Units' digit of \(7^{37}\) = Units' digit of \(7^1\) = 7

Q3. Find Units’ digit of \(7^{52}\)?

Sol: 52 divided by 4 gives 0 remainder
=> Units' digit of \(7^{52}\) = Units' digit of \(7^4\) = 1

Q4. Find Units’ digit 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
=> Units' digit of \(7^{80a + 51}\) = Units' digit of \(7^3\) = 3

Q5. Find Units’ digit of \(1297^{2041}\)?

Sol: Units' digit of power of any number = Units' digit of power of the units' digit of that number
=> Units’ digit of \(1297^{2041}\) = Units’ digit of \(7^{2041}\)
=> Remainder of 2041 divided by 4 = Remainder of last two digits by 4

Watch this video to Master Divisibility Rules

=> Remainder of 41 by 4 = 1
=> Units' digit of \(1297^{2041}\) = Units' digit of \(7^1\) = 7

Hope it helps!
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Attachment:
Units Digit of Power of 8.pdf [185.75 KiB]
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How to Solve: Units' Digit of Power of 8


Hi All,

I have posted a video on YouTube to discuss Units' Digit of Power of 8



Attached pdf of this Article as SPOILER at the top! Happy learning! :)

Following is Covered in the Video

Theory of Units' Digit of Power of 8
    ⁍ Find Units’ digit of \(8^{91}\) ?
    ⁍ Find Units’ digit of \(8^{57}\) ?
    ⁍ Find Units’ digit of \(8^{88}\) ?
    ⁍ Find Units’ digit of \(8^{40a + 41}\) (given that a is a positive integer)?
    ⁍ Find Units’ digit of \(1738^{8979}\) ?


Theory of Units' Digit of Power of 8

• To find units' digit of any positive integer power of 8


We need to find the cycle of units' digit of power of 8
\(8^1\) units’ digit is 8
\(8^2\) units’ digit is 4
\(8^3\) units’ digit is 2
\(8^4\) units’ digit is 6
\(8^5\) units’ digit is 8
\(8^6\) units’ digit is 4
\(8^7\) units’ digit is 2
\(8^8\) units’ digit is 6

=> The power repeats after every \(4^{th}\) power
=> Cycle of units' digit of power of 8 = 4
=> We need to divide the power by 4 and check the remainder
=> Units' digit will be same as Units' digit of \(8^{Remainder}\)

NOTE: If Remainder is 0 then units' digit = units' digit of \(8^{Cycle}\) = units' digit of \(8^{4}\) = 1

Q1. Find Units’ digit of \(8^{81}\)?

Sol: We need to divided the power (81) by 4 and get the remainder
81 divided by 4 gives 1 remainder
=> Units' digit of \(8^{81}\) = Units' digit of \(8^1\) = 8

Q2. Find Units’ digit of \(8^{57}\)?

Sol: 57 divided by 4 gives 1 remainder
=> Units' digit of \(8^{57}\) = Units' digit of \(8^1\) = 8

Q3. Find Units’ digit of \(8^{88}\)?

Sol: 88 divided by 4 gives 0 remainder
=> Units' digit of \(8^{88}\) = Units' digit of \(8^4\) = 6

Q4. Find Units’ digit of \(8^{40a + 41}\) (given that a is a positive integer)?

Sol: Remainder of 40a + 41 divided by 4 = Remainder of 40a by 4 + Remainder of 41 by 4
= 0 + 1 = 1
=> Units' digit of \(8^{40a + 41}\) = Units' digit of \(8^1\) = 8

Q5. Find Units’ digit of \(1738^{8979}\)?

Sol: Units' digit of power of any number = Units' digit of power of the units' digit of that number
=> Units’ digit of \(1738^{8979}\) = Units’ digit of \(8^{8979}\)
=> Remainder of 8979 divided by 4 = Remainder of last two digits by 4

Watch this video to Master Divisibility Rules

=> Remainder of 79 by 4 = 3
=> Units' digit of \(1738^{8979}\) = Units' digit of \(8^3\) = 2

Hope it helps!­
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Attachment:
Units Digit of Power of 9.pdf [158.67 KiB]
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How to Solve: Units' Digit of Power of 9


Hi All,

I have posted a video on YouTube to discuss Units' Digit of Power of 9



Attached pdf of this Article as SPOILER at the top! Happy learning! :)

Following is Covered in the Video

Theory of Units' Digit of Power of 9
    ⁍ Find Units’ digit of \(9^{81}\) ?
    ⁍ Find Units’ digit of \(9^{53}\) ?
    ⁍ Find Units’ digit of \(9^{68}\) ?
    ⁍ Find Units’ digit of \(9^{60x + 61}\) (given that x is a positive integer)?
    ⁍ Find Units’ digit of \(13259^{1279}\) ?


Theory of Units' Digit of Power of 9

• To find units' digit of any positive integer power of 9


We need to find the cycle of units' digit of power of 9
\(9^1\) units’ digit is 9
\(9^2\) units’ digit is 1
\(9^3\) units’ digit is 9
\(9^9\) units’ digit is 1

=> The power repeats after every \(2^{nd}\) power
=> Cycle of units' digit of power of 9 = 2

=> Units' digit of odd power of 9 = 9
=> Units' digit of even power of 9 = 1

Q1. Find Units’ digit of \(9^{81}\)?

Sol: 81 is Odd
=> Units' digit of \(9^{81}\) = 9

Q2. Find Units’ digit of \(9^{53}\)?

Sol: 53 is Odd
=> Units' digit of \(9^{53}\) = 9

Q3. Find Units’ digit of \(9^{68}\)?

Sol: 68 is Even
=> Units' digit of \(9^{Even}\) = 1

Q9. Find Units’ digit of \(9^{60x + 61}\) (given that x is a positive integer)?

Sol: 60x + 61 = Even + Odd = Odd
=> Units' digit of \(9^{60x + 61}\) = 9

Q5. Find Units’ digit of \(13259^{1279}\) ?

Sol: Units' digit of power of any number = Units' digit of power of the units' digit of that number
=> Units’ digit of \(13259^{1279}\) = Units’ digit of \(9^{1279}\)
=> 1279 is Odd
=> Units' digit of \(13259^{1279}\) = Units, digit of \(9^{1279}\)= 9

Hope it helps!
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Attachment:
Units Digit of Product of Numbers.pdf [153.27 KiB]
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How to Solve: Units' Digit of Product of Numbers


Hi All,

I have posted a video on YouTube to discuss Units' Digit of Product of Numbers



Attached pdf of this Article as SPOILER at the top! Happy learning! :)

Following is Covered in the Video

Theory of Units' Digit of Product of Numbers
    ⁍ Find Units’ digit of 23 * 34 ?
    ⁍ Find Units’ digit of 236 * 347 * 12468 ?
    ⁍ Find Units’ digit of 2248 * 285 * 980459 ?
    ⁍ Find Units’ digit of 480 * 285745644 * 980459 * 3213123 ?


Theory of Units' Digit of Product of Numbers

• To determine the units' digit of a product of numbers, extract the units' digit of each number, multiply them together, and repeat this process until you obtain the units' digit of the final result.

Q1. Find Units’ digit of 23 * 34 ?

Sol: We will take the units' digit of each number and multiply them together
Units' digit of 23 = 3 and Units' digit of 34 is 4
=> 3 * 4 = 12
=> Units' digit of 23 * 34 = 2

Q2. Find Units’ digit of 236 * 347 * 12468 ?

Sol: 6 * 7 * 8 = 42 * 8 = ... 6
=> Units' digit of 236 * 347 * 12468 = 6

Q3. Find Units’ digit of 2248 * 285 * 980459 ?

Sol: 8 * 5 * 9 = 40 * 9 = ...0
=> Units' digit of 2248 * 285 * 980459 = 0

Q4. Find Units’ digit of 480 * 285745644 * 980459 * 3213123 ?

Sol: 0 * .....
Whenever we have 0 multiplied by any number then units' digit will always be 0
=> Units' digit of 480 * 285745644 * 980459 * 3213123 = 0

MASTER Units' Digit of Exponents by going through this post.

Hope it helps!
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Attachment:
Units Digit of Product of Exponents.pdf [229.8 KiB]
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How to Solve: Units’ Digit of Product of Exponents


Hi All,

I have posted a video on YouTube to discuss Units’ Digit of Product of Exponents



Attached pdf of this Article as SPOILER at the top! Happy learning! :)

Following is Covered in the Video

Theory of Units’ Digit of Product of Exponents
    ⁍ Find Units’ digit of \(4^{23} * 3^{46}\) ?
    ⁍ Find Units’ digit of \(2^{34} * 3^{45}\) ?
    ⁍ Find Units’ digit of \(1452^{48} * 25463^{123} * 798^{241}\) ?
    ⁍ Find Units’ digit of \(145^{2367} * 156^{10987}?\)
    ⁍ Find Units’ digit of \(224^{12987} * 28569^{1879}\)?
    ⁍ Cyclicity of Units’ digit of numbers ( 1 to 10 )


Theory of Units’ Digit of Product of Exponents

To find the units' digit of a product of exponents, follow these steps:
    1. Examine if the exponents can be rearranged to consolidate them into a single exponent and then find the units’ digit using cyclicity of the number.
    2. If rearrangement is not possible, determine the units' digit of each exponent and multiply them to obtain the units' digit of the entire expression.


Q1. Find Units’ digit of \(4^{23} * 3^{46}\) ?

Sol: \(4^{23}\) = \((2^2)^{23}\) = \(2^{46}\)
=> \(4^{23} * 3^{46}\) = \(2^{46} * 3^{46}\) = \((2*3)^{46}\) = \(6^{46}\)
Cyclicity of units' digit of power of 6 is 1
[Go through this post to MASTER Cyclicity of Units' digit of numbers from 2-9 ]

=> Units’ digit of \(4^{23} * 3^{46}\) = 6

Q2. Find Units’ digit of \(2^{34} * 3^{45}\) ?

Sol: Cyclicity of units' digit of power of 2 and 3 is 4
Units' digit of \(2^{34}\) = Units' digit of \(2^{2}\) (as remainder of 34 by 4 is 2) = 4
Units' digit of \(3^{45}\) = Units' digit of \(3^{1}\) (as remainder of 45 by 4 is 2) = 1
=> Units’ digit of \(2^{34} * 3^{45}\) = 4 * 1 = 4

Q3. Find Units’ digit of \(1452^{48} * 25463^{123} * 798^{241}\) ?

Sol: Whenever we have to find units digit of power of a big number then we just need to focus on the units' digit of the number and take its power.
=> Units' digit of \(1452^{48} * 25463^{123} * 798^{241}\) = Units' digit of \(2^{48} * 3^{123} * 8^{241}\)
Cyclicity of units' digit of power of 2, 3 and 8 is 4
Units' digit of \(2^{48}\) = Units' digit of \(2^{4}\) (as remainder of 48 by 4 is 0 so we take unit's digit of the power of the cycle, which is 4) = 6
Units' digit of \(3^{123}\) = Units' digit of \(3^{3}\) (as remainder of 123 by 4 is 3) = 7
Units' digit of \(8^{241}\) = Units' digit of \(8^{1}\) (as remainder of 241 by 4 is 1) = 8
=> Units’ digit of \(1452^{48} * 25463^{123} * 798^{241}\) = 6 * 7 * 8 = 42 * 8 = ...6

Q4. Find Units’ digit of \(145^{2367} * 156^{10987}\)?

Sol: Units' digit of \(145^{2367} * 156^{10987}\) = Units' digit of \(5^{2367} * 6^{10987}\)
=> Both 5 and 6 have a cycle of 1
=> Units' digit of \(145^{2367} * 156^{10987}\) = 5 * 6 = _0

Q5. Find Units’ digit of \(224^{12987} * 28569^{1879}\)?

Sol: Units' digit of \(224^{12987} * 28569^{1879}\) = Units' digit of \(4^{12987} * 9^{1879}\)
Cyclicity of units' digit of power of 4 and 9 is 2
=> Units' digit of \(4^{12987}\) = Units' digit of \(4^1\) = 4
=> Units' digit of \(9^{1879}\) = Units' digit of \(9^1\) = 9
=> Units’ digit of \(224^{12987} * 28569^{1879}\) = 4 * 9 = 6

Cyclicity of Units’ digit of numbers ( 1 to 10 )

Attachment:
Cyclicity.jpg
Cyclicity.jpg [ 127.71 KiB | Viewed 5662 times ]

MASTER Units' Digit of Exponents by going through this post.

Hope it helps!
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Last Two Digits of Exponents Ending with 1


Hi All,

I have posted a video on YouTube to discuss Last Two Digits of Exponents Ending with 1



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 \(131^{345}\) ?
    ⁍ Find Last two digits of \(781^{984}\) ?
    ⁍ Find Last two digits of \(15671^{379}\) ?


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 \(131^{345}\) ?

Sol: Base = 131
Exponent = 345

=> Units' digit = 1
=> Tens' digit = 3 * 5 [131 * 345]
= 5
=> Last two digits = 51

Q2. Find Last two digits of \(781^{984} \)?

Sol: Base = 781
Exponent = 984

=> Units' digit = 1
=> Tens' digit = 8 * 4 [781 * 984]
= 2
=> Last two digits = 21

Q3. Find Last two digits of \(15671^{379}\) ?

Sol: Base = 15671
Exponent = 379

=> Units' digit = 1
=> Tens' digit = 7 * 9 [15671 * 379]
= 3
=> Last two digits = 31

Link to Theory for Last Two digits of exponents here.

Link to Theory for Units' digit of exponents here.

Hope it helps!­
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Last Two Digits of Exponents Ending with 2


Hi All,

I have posted a video on YouTube to discuss Last Two Digits of Exponents Ending with 2



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 2
    ⁍ Find Last two digits of \(2^{4274}\) ?
    ⁍ Find Last two digits of \(842^{9802}\) ?


Theory of Last Two Digits of Numbers Ending with 2

    • Express the Number as \((2^{10})^{Power}\) * \(2^{Smaller Power}\)
    • Now we know that \(2^{10}\) = 1024 and we have expressed the number \(1024^{Power}\)
    • \(24^{Odd Power}\) will have last two digits as 24
    • \(24^{Even Power}\) will have last two digits as 76
    • If we have power of power then we can use last two digits of \(76^{Any Positive Integer}\) is 76


Q1. Find Last two digits of \(2^{4274}\) ?

Sol: \(2^{4274}\) = \(2^{4270 + 4}\) = \(2^{10 * 427} * 2^4\)
= \((2^{10})^{ 427} * 16\) = \(1024^{427} * 16\)
= \(1024^{Odd} * 16\)
=> Last two digits 24 * 16
=> Last two digits = 84

Q2. Find Last two digits of \(842^{9802}\) ?

Sol: \(842^{9802}\) = \((421 * 2)^{9802}\)
= \((421)^{9802} * 2^{9800 + 2}\)
=> Last two digits = 41 * Last two digits of \((2^{10})^{980} * 2^2\) [Watch this video to learn about How to Find Last two digits of Exponents ending with 1]
=> Last two digits = 41 * Last two digits of \(1024^{Even} * 4\)
=> Last two digits = 41 * Last two digits of 76 * 4
=> Last two digits = 64

Link to Theory for Last Two digits of exponents here.

Link to Theory for Units' digit of exponents here.

Hope it helps!­
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Last Two Digits of Exponents Ending with 5


Hi All,

I have posted a video on YouTube to discuss Last Two Digits of Exponents Ending with 5



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 5
    ⁍ Find Last two digits of \(735^{637}\) ?
    ⁍ Find Last two digits of \(785^{989}\) ?
    ⁍ Find Last two digits of \(15635^{372}\) ?


Theory of Last Two Digits of Numbers Ending with 5

If the exponents is of the format:
    • \((O5)^O\), where O is Odd, then last two digits = 75
    • Otherwise, last two digits = 25 (i.e. \((O5)^E\) , \((E5)^O\), \((E5)^E\) , where O is Odd and E is Even)


Q1. Find Last two digits of \(735^{637}\) ?

Sol: Number before 5 is Odd (735)
Exponent = Odd (637)
=> \((O5)^O\) case
=> Last two digits = 75

Q2. Find Last two digits of \(785^{989}\) ?

Sol: Number before 5 is Even (785)
Exponent = Odd (989)
=> \((E5)^O\) case
=> Last two digits = 25

Q3. Find Last two digits of \(15635^{372}\) ?

Sol: Number before 5 is Odd(15635)
Exponent = Even (373)
=> \((O5)^E\) case
=> Last two digits = 25

Link to Theory for Last Two digits of exponents here.

Link to Theory for Units' digit of exponents here.

Hope it helps!­
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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 \(131^{345}\) ?

Sol: Base = 131
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 \(3^{241}\) ?

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

Q3. Find Last two digits of \(783^{402}\) ?

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

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 \(7^{282}\) ?

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

Q5. Find Last two digits of \(847^{422}\) ?

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

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 \(9^{243}\) ?

Sol: \(9^{242 + 1}\)
= \(9^{2*121} * 9^1\)
= \((9^2)^{121} * 9\)
= \((81)^{121} * 9\)
= 81 * 9 = 29
=> Last two digits = 29

Q7. Find Last two digits of \(1269^{436}\) ?

Sol: \((141*9)^{436}\)
= \((141)^{436} * 9^{2*218}\)
= 41 * \((9^2)^{218}\)
= 41 * \((81)^{218}\)
= 41 * 41
= 81
=> Last two digits = 81


Link to Theory for Units' digit of exponents here.

Hope it helps!­
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Last Two Digits of Exponents Ending with 2, 4, 6, 8


Hi All,

I have posted a video on YouTube to discuss Last Two Digits of Exponents Ending with 2, 4, 6, 8



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 2
    ⁍ Find Last two digits of \(2^{4274}\) ?
    ⁍ Theory of Last Two Digits of Numbers Ending with 4
    ⁍ Find Last two digits of \(4^{501}\) ?
    ⁍ Find Last two digits of \(1684^{8101}\) ?
    ⁍ Theory of Last Two Digits of Numbers Ending with 6
    ⁍ Find Last two digits of \(6^{321}\) ?
    ⁍ Find Last two digits of \(486^{422}\) ?
    ⁍ Theory of Last Two Digits of Numbers Ending with 8
    ⁍ Find Last two digits of \(8^{201}\) ?
    ⁍ Find Last two digits of \(1768^{821}\) ?


Theory of Last Two Digits of Numbers Ending with 2

    • Express the Number as \((2^{10})^{Power}\) * \(2^{Smaller Power}\)
    • Now we know that \(2^{10}\) = 1024 and we have expressed the number \(1024^{Power}\)
    • \(24^{Odd Power}\) will have last two digits as 24
    • \(24^{Even Power}\) will have last two digits as 76
    • If we have power of power then we can use last two digits of \(76^{Any Positive Integer}\) is 76


Q1. Find Last two digits of \(2^{4274}\) ?

Sol: \(2^{4274}\) = \(2^{4270 + 4}\) = \(2^{10 * 427} * 2^4\)
= \((2^{10})^{ 427} * 16\) = \(1024^{427} * 16\)
= \(1024^{Odd} * 16\)
=> Last two digits 24 * 16
=> Last two digits = 84

Theory of Last Two Digits of Numbers Ending with 4

    • We will convert the number from \(4^{SomePower}\) to \(2^{2*SomePower}\)
    • Apply the rule for power of 2 as described above


Q2. Find Last two digits of \(4^{501}\) ?

Sol: \(4^{501}\) = \(2^{2*501}\)
= \(2^{1002}\)
= \(2^{1000 + 2}\)
= \(2^{10*100} * 2^2\)
= \((2^{10})^{100} * 4\)
= \((1024)^{Even} * 4\)
= \(76 * 4\)
= 04
=> Last two digits = 04

Q3. Find Last two digits of \(1684^{8101}\) ?

Sol: \((421*4)^{8101}\)
= \((421)^{8101} * 2^{2*8101}\)
= 21 * \(2^{16202}\)
= 21 * \(2^{16200 + 2}\)
= 21 * \(2^{10*1620}* 2^2\)
= 21 * \((2^{10})^{1620}* 4\)
= 21 * \((1024)^{Even}* 4\)
= 84 * 76
= 84
=> Last two digits = 84

Theory of Last Two Digits of Numbers Ending with 6

    • We will convert the number from \(6^{SomePower}\) to \(2^{SomePower} * 3^{SomePower}\)
    • Apply the rule for power of 2 and 3 to get the answer
[Use this link to learn how to find last two digits of numbers ending with 3]

Q4. Find Last two digits of \(6^{321}\) ?

Sol: \(6^{321}\) = \(2^{321} * 3^{321}\)
= \(2^{320 + 1} * 3^{320 + 1}\)
= \(2^{10*32} * 2^1 * 3^{4*80} * 3^1\)
= \((2^{10})^{32} * 2 * (3^{4})^{80} * 3\)
= \((1024)^{Even} * 2 * (81)^{80} * 3\)
= \(76 * 6 * 01\)
= 56
=> Last two digits = 56

Q5. Find Last two digits of \(486^{422}\) ?

Sol: \((243*2)^{422}\)
= \((3^5)^{422} * 2^{420 + 2}\)
= \(3^{5 *422} * 2^{10*42} * 2^2\)
= \(3^{2110} * 1024^{Even} * 4\)
= \(3^{2108+2} * 76 * 4\)
= \(3^{4 * 527} * 3^2 * 04\)
= \(81^{527} * 9 * 04\)
= \(61 * 36\)
= 96
=> Last two digits = 96

Theory of Last Two Digits of Numbers Ending with 8

    • We will convert the number from \(8^{SomePower}\) to \(2^{3*SomePower}\)
    • Apply the rule for power of 2 as described above


Q6. Find Last two digits of \(8^{201}\) ?

Sol: \(8^{201}\) = \(2^{3*201}\)
= \(2^{603}\)
= \(2^{600 + 3}\)
= \(2^{10*60} * 2^3\)
= \((2^{10})^{60} * 8\)
= \((1024)^{Even} * 8\)
= \(76 * 8\)
= 08
=> Last two digits = 08

Q7. Find Last two digits of \(1768^{821}\) ?

Sol: \((221*8)^{821}\)
= \((221)^{821} * 2^{3*821}\)
= 21 * \(2^{2463}\)
= 21 * \(2^{2460 + 3}\)
= 21 * \(2^{10*246}* 2^3\)
= 21 * \((2^{10})^{246}* 8\)
= 21 * \((1024)^{Even}* 8\)
= 68 * 76
= 68
=> Last two digits = 68


Link to Theory for Last Two digits of exponents here.

Link to Theory for Units' digit of exponents here.

Hope it helps!­
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­GMAT Quant Theory Master for Post and Videos on Finding Last Two Digits of Exponents by BrushMyQuant
(PDFs can be found in individual links given as GmatClub Article Link)





Last Two Digits of Exponents Ending with 1, 3, 7 and 9





Last Two Digits of Exponents Ending with 2, 4, 5, 6 and 8


Link for Finding Units Digits of Numbers here.

Hope it helps!­
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Last Two Digit of Power of 7.pdf [196.47 KiB]
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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



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
\(7^5\) 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 \(7^{93}\)?

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

Q2. Find Last two digits of \(7^{1529}\)?

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

Hope it helps!

Link to Theory for Last Two digits of exponents here.

Link to Theory for Units' digit of exponents here.
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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 \(6^{71}\)?

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

Q2. Find Last two digits of \(6^{12517}\)?

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

Hope it helps!

Link to Theory for Last Two digits of exponents here.

Link to Theory for Units' digit of exponents here.
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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 \(11^{128}\)?
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

Q2. Find Last two digits of \(11^{15342}\)?
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

Hope it helps!

Link to Theory for Last Two digits of exponents here.

Link to Theory for Units' digit of exponents here.
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