# 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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• ##### Keshav

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

I get it now, Thanks

• ##### Maharshi

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.

• ##### Maharshi

Thank you very much.. You da real MVP 😉

• ##### Benedict

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

• ##### Sujith Patnaik

Thanks Benedict :). I updated it.

• ##### Shweta Balachandran

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

Yay! Thanks a lot. You’re awesome..

• ##### Subodh

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

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

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

• ##### raj

Solve (2734) power 123 the epuation

• ##### Dikshika

Last two digits of 37^63?

• ##### ANTO

287 HOW TO FIND THE LAST DIGIT NUMBER