Last Two Digits of Large Number

Here, we will see how to find the last two digits of a large number. We will first understand how to find the last two digits of a product and  then we will try to find the last two digits ( i.e digits in tens and units place) of a large number in the form ${x^y}$.

At the end of this post, we have a quiz for you that will test your understanding of the techniques to find the last two digits of large numbers in the form $x^y$

Video :

The video contains on explanation on how to find the last two digits of a number raised to power based on the units digit of the number.

Understanding the basics – Last two digits of a product

Given a number say 1439, the last two digits of this number is 3 and 9, which is straight forward. Now, how do we find the last two digits in the product of 1439 x 2786? Just follow the steps given below :

In the product of two numbers say A and B, (in our case A is 1439 and B is 2786)

If a and b, respectively represent the digits in the ten’s place and one’s place of A and similarly c and d respectively represent the digits in the ten’s place and one’s place of B, then

  1. The units digit in the product of b and d is the units digit in AxB and the digit in ten’s place(if any) becomes the carry. i.e in 1439×2786, multiply 9 with 6 which gives 54. Here 4 forms the units digit of 439×786 and 5 goes as carry to Step 2.
  2. Add the product of a and d with the product of c and b. i.e in 1439×2786 => 3×6 + 9×8 = 90
  3. If a carry is generated in Step1, add that with the result obtained in Step 2. i.e 5 + 90 = 95
  4. The units digit in the result obtained in Step 3 forms the tens digit in the product of A and B. i.e the units digit in 95 which is 5 becomes the tens digit of 1439×2786.

Hence the last two digits of 1439×2786 is 5 and 4.

How to find the Last Two Digits of Number raised to Power

Let the number be in the form ${x^y}$. Based on the value of units digit in the base i.e x, we have four cases

Case 1: Units digit in x is 1

If x ends in 1, then x raised to y, ends in 1 and its tens digit is obtained by multiplying the tens digit in x with the units digit in y.

Example 1: Find the last two digits of ${91^{246}}$

Since the base 91 ends in 1, ${ 91^{246}}$ ends in 1 and the tens place digit is obtained from the units digit in 9×6 which is 4. Hence the last two digits of ${ 91^{246}}$ are 4 and 1.

Case 2: Units digit in x is 3, 7 or 9

In this case we will convert the base so that it ends in 1, after which we can use Case 1 to calculate units and tens place digits. i.e

When x ends in 9 ${(..9)^{y}}$

Raise the base by 2 and divide the exponent by 2 => ${(..9^2)^{y/2}}$

Number ending in 9 raised to 2 ends in 1 => ${(..1)^{y/2}}$

Since the base now ends in 1, Tens digit and Units digit is calculated using the steps in Case 1.

Example 2: Find the last two digits of ${(79)^{142}}$ => ${(79^2)^{142/2}}$ => ${(..41)^{71}}$

Now, units digit of a number ending in 41 to the power of 71 is 1 and its tens digit is obtained by multiplying 4 and 1 which is 4.

Hence, the last two digits of ${(79)^{142}}$ are 4 and 1.

Example 3: Find the last two digits of ${(79)^{143}}$ => $(79)^{142}$ x ${(79)^1}$

$(79)^{142}$ ends in 41 (From previous example) and ${(79)^1}$ ends in 79. Hence, the product of numbers ending in 41 and 79 ends in 39, which implies the last two digits of ${(79)^{143}}$ are 3 and 9.

When x ends in 3 ${(..3)^{y}}$

Raise the base by 4 and divide the exponent by 4 => ${(..3^4)^{y/4}}$

Number ending in 3 raised to 4 ends in 1 => ${(..1)^{y/4}}$

Since the base now ends in 1, Tens digit and Units digit is calculated using the steps in Case 1.

Example 4: Find the last two digits of ${(43)^{76}}$ => ${(43^4)^{76/4}}$ => ${(..01)^{19}}$

Now, units digit of a number ending in 01 to the power of 19 is 1 and its tens digit is obtained by multiplying 0 and 9 which is 0.

Hence, the last two digits of ${(43)^{76}}$ are 0 and 1.

When x ends in 7 ${(..7)^{y}}$

Raise the base by 4 and divide the exponent by 4 => ${(..7^4)^{y/4}}$

Number ending in 7 raised to 4 ends in 1 => ${(..1)^{y/4}}$

Since the base now ends in 1, Tens digit and Units digit is calculated using the steps in Case 1.

Example 5: Find the last two digits of ${(17)^{256}}$ => ${(17^4)^{256/4}}$ => ${(..21)^{44}}$

Units digit of a number ending in 21 to the power of 44 is 1 and its tens digit is obtained by multiplying 2 and 4 which is 8.

Hence, the last two digits of ${(17)^{256}}$ are 8 and 1.

Case 3: Units digit in x is 2, 4, 6 or 8

If x ends in 2, 4, 6, 0r 8, we can find the last two digits of the number raised to power with the help of the following points :

  • ${(2)^{10}}$ ends in 24
  • ${(24)^{odd\ number}}$ ends in 24
  • ${(24)^{even\ number}}$ ends in 76
  • ${(76)^{number}}$ ends in 76

Example 6: Find the last two digits of 2 raised to 1056

${(2)^{1056}}$ => $(2^{10})^{105}$ x ${(2)^{6}}$

Here, 2 raised to 10 ends in 24 and 24 raised 105, which is an odd number, ends in 24. Also 2 raised to 6 ends in 64. Hence the last two digits of the product of the numbers ending in 24 and 64 are 3 and 6.

Case 4: Units digit in x is 5

  • If the digit in the tens place is odd and the exponent y is odd, then the number ends in 75.
  • If the digit in the tens place is odd and the exponent y is even, then the number ends in 25.
  • If the digit in the tens place is even and the exponent y is odd, then the number ends in 25.
  • If the digit in the tens place is even and the exponent y is even, then the number ends in 25 .

Hence when the exponent and the digit in the tens place of the base are odd, the number raised to power ends 75, in other cases it ends in 25.

Example 7: Find the last two digits of ${(65)^{243}}$

Since the digit in the tens place of the base is even and the exponent is odd, last two digits of ${(65)^{243}}$  are 2 and 5.

Example 8: Find the last two digits of ${(135)^{1091}}$

Since the digit in the tens place of the base is odd and the exponent is odd, ${(65)^{243}}$  ends in 75.


Quiz : Test your Understanding

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Comments

  • Keshav
    Reply

    regarding the shortcut mentioned about finding last two places of a number ending with 5, raised to odd power,

    it will always end up in 25.

    but u have mentioned as 75.

    kindly clarify…!!

    • Sujith Patnaik

      Hi Keshav,

      In addition to the power, you also need to check whether the digit in ten’s place is odd or even –
      The last two digits of a number ending in 5, raised to power will be 75 when the digit in the tens place is odd and the power is odd.

      Example : $35^3$ = 42875

  • Keshav
    Reply

    I get it now, Thanks

  • Maharshi
    Reply

    13^17 + 17^13 is divisible by 10?
    A GRE question.
    Thank you.

    • Sujith Patnaik

      It is divisible by 10.
      Units digit of 13^17 is 3 and units digit of 17^13 is 7.
      Sum of the units digits, 3 and 7 is 10. Hence 13^17 + 17^13 ends in 0 which implies that it is divisible by 10.

      To find the units digit Click here

    • Maharshi

      Thank you very much.. You da real MVP 😉

  • Benedict
    Reply

    In the second question of your quiz. Find the last two digits of 19^347 – 22^120. You used 19^346 in the solution.

  • Shweta Balachandran
    Reply

    How would you find the last two digits of (367) ^ 1054? As for the method mentioned, it doesn’t work since 1054 is not divisible by 4.

    • Sujith Patnaik

      (367) ^ 1054 = ((367) ^ 1052) x (367) ^ 2
      = ((367^4) ^ 263) x (..89) [Last two digits of 367^2 = 89]
      =(21^263) x 89 [Last two digits of ..67^4 = 21]
      = 61 x 89
      = ..29

  • Shweta balachandran
    Reply

    Yay! Thanks a lot. You’re awesome..

  • Subodh
    Reply

    How to find remainder when 29^202 divided by 13?

    • Sujith Patnaik

      Approach 1: Using the basic concept of remainders

      $\frac{29^{202}}{13}$ =>

      29 leaves a remainder 3 when divided by 13. Hence,

      $\frac{29^{202}}{13}$ => $\frac{3^{202}}{13}$

      $\frac{3^{202}}{13}$ = $\frac{3^{201} \text{ x } 3}{13}$

      = $\frac{(3^{3})^{67} \text{ x } 3}{13}$

      $3^3$ leaves remainder 1 when divided by 13. Hence

      Remander of $\frac{(3^{3})^{67} \text{ x } 3}{13}$ =>Remainder of $\frac{(1)^{67} \text{ x } 3}{13}$ = 3

      Approach 2: Using the Euler’s theorem described in this link -www.justquant.com/numbertheory/remainders-using-fermats-and-eulers-theorem/

      $\frac{29^{202}}{13}$ =>

      Since 29 and 13 are co primes, we can apply Euler’s theorem.

      -> Calculate the Euler’s totient function of 13.
      $=> \Phi(13) = 13(1 – \frac{1}{13})$ = 12

      -> Divide the exponent 202 by $\Phi(13)$ to find remainder y.
      $ y = $Remainder$(\frac{202}{\Phi(13)})$ = Remainder$(\frac{202}{12})$ => $10$

      -> Find the remainder of 3 to the power of $y$ when divided by 13.
      Remainder$(\frac{3^{y}}{13})$ = Remainder$(\frac{3^{10}}{13})$ = Remainder $(\frac{(3^3)^3 \text{ x } 3}{13})$ = Remainder$(\frac{1 \text{ x } 3}{13})$ = 3

      Therefore,Remainder of $\frac{29^{202}}{13}$ = $3$

  • Abhishek
    Reply

    How to find the remainder when (164)^359 is divided by 100.

    • Sujith Patnaik

      164$^{359}$ = 41$^{359}$ x 4$^{359}$

      Last two digits of 41$^{359}$ => 61 [Using case 1]

      Last two digits of 4$^{359}$
      => Last two digits of 2$^{718}$ = 44 [Using case 3]

      Hence last two digits of 164$^{359}$ => Last two digits of 61 x 44 => 84

  • Abhishek
    Reply

    What is the remainder when 5^119 is divided by 59?

  • raj
    Reply

    Solve (2734) power 123 the epuation

  • Dikshika
    Reply

    Last two digits of 37^63?

  • ANTO
    Reply

    287 HOW TO FIND THE LAST DIGIT NUMBER

  • soniya
    Reply

    what is the unit digit of 1234567^7654321 ?

6 + 4 =