What are the last two digits of the number 7 raise to 45?

Here, we will see how to find the last two digits of a large number. In the previous post we tried to understand calculating the units digits of large numbers like $2014^{2012}$, $1453^{71}$ etc. In this post,

• 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, take the quiz to 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

Last two digits of a number is basically the tens place and units place digit of that number.  So given a number say 1439, the last two digits of this number are 3 and 9, which is pretty straight forward. Now, how do we find the last two digits in the product of 1439 x 2786? One feasible approach is using the vertical and cross wise technique of multiplication.

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. Units digit of A x B is given by the units digit in the product of b and d. If the product of b and d results in more than 1 digit, the excess digit will be carried over to the left. i.e in 1439 x 2786, multiply 9 and 6 which gives 54. Here 4 forms the units digit of 1439 x 2786 and 5 goes as carry to Step 2.
2. In this step, we cross multiply a and d & c and b, and then add the resulting products. i.e in 1439 x 2786 => 3×6 + 9×8 = 90
3. If a carry is generated in Step 1, 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 x 2786.

Hence the last two digits of 1439 x 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. Using the vertical and cross-wise multiplication technique 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

• 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.
• Or 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

What Next?

Can you calculate the highest power of  15 in 24! or number of zeros at the end of 1000!? Click on the post below to learn more about highest power of numbers in a factorial.

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What are the last two digits of the number 7^45 [#permalink]

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What are the last two digits of the number 7^45A 07B 23C 49D 43

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Re: What are the last two digits of the number 7^45 [#permalink]

The remainder when \(7^{45}\) is divided by 100 is the last 2 digits of \(7^{45}\).\(7^2\)=49= 49 Mod 100\(7^4\)= 2401= (01) mod 100\(7^{4k}\)= 01 MOD 100\(7^{4k+1}\)= (01*7) MOD 100= 07 MOD 10045= 4*11+1, hence \(7^{45}\)= 07 MOD 100A

AbdulMalikVT wrote:

What are the last two digits of the number 7^45A 07B 23C 49D 43

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Re: What are the last two digits of the number 7^45 [#permalink]

AbdulMalikVT wrote:

What are the last two digits of the number 7^45A 07B 23C 49D 43

E 27

so for power 7^45 ; 7^4x+1 ; unit digit will be 3 and tens digit will be same as 7^4 ; 2401

Re: What are the last two digits of the number 7^45 [#permalink]

AbdulMalikVT wrote:

What are the last two digits of the number 7^45A 07B 23C 49D 43

E 27

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Re: What are the last two digits of the number 7^45 [#permalink]

To find the last two digits we essentially need to find the remainder when divided by 100. now, 7 follows cyclicity of 4 i.e. after the fourth power the unit digit gets repeated. in this case we need last two digits.

7^4 = 2401. now last two digits are 01

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Re: What are the last two digits of the number 7^45 [#permalink]

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