**Last Two digits:** The objective of this article is to understand the concept and approach to solve questions like:

Find the last two digits of

- \({2^{234}}\)
- \({31^{123}}\)
- \({73^{35}}\)

There are two-three shortcut approaches to solve the above problems. But before we go and apply the same, we need to understand the following basic points.

### Finding the last two digits of a square of any number:

Let’s take an example; say we need to find the last two digits of \({78^2}\). *The prerequisite is to memorise the square of the first 25 natural numbers.*

Now, number 78 is close to 100 and the deviation of 78 from 100 is 22 (100-78). So we can write \({78^2} = {\left( {100 – 22} \right)^2} = {100^2} – 2 \times 100 \times 22 + {22^2} = 10000 – 4400 + 484\)

Now observe that the last two digits of the first two terms are 00, 00, and 84.

So the last two digits of \({78^2} \equiv \) last two digits of \({\left( {100 – 22} \right)^2} \equiv \) last two digits of \({22^2}\) Similarly, the last two digits of \({61^2} \equiv {\left( {50 + 11} \right)^2} \equiv {11^2} \equiv 21\)

From the above examples, we can deduce that the if the number is close to 100 or 50, the last two digits of the square of the number is equal to the ** last two digits of the square of the deviation of the number** from the base 100 or 50 (whichever the number if close to).

### Finding the tens place digit of any number with unit digit equal to 1

Say we need to find the last two digits of \({31^{76}}\).

The trick to finding the second last digit of any number ending with 1 is the **UNIT DIGIT** of the product of the unit digit of the power and the ten’s digit of the base.

It is clear that the unit digit of the given number is 1 and we have to identify the ten’s place digit of the same.

In \({31^{76}}\), the base is 31 (ten’s place is 3) and power is 76 (the unit digit is 6) . So the second last (ten’s digit) of \({31^{76}}\) is Unit digit of \(3 \times 6 = 18\) or 8.

Therefore, the last two digits of \({31^{76}}\) are 81.

**Note:** To verify the above result, we can use the concept of binomial expansion. I leave the same to the readers to do.

#### Try a few more examples

Example: Find the tens place digit of \({41^{38}}\), \({71^{87}}\), \({91^{94}}\)

The correct answers are 2, 9, and 6 respectively. I hope you got all right. If not, then please revisit the concept before moving ahead.

### Finding last two digits of odd numbers ending with 3, 7 or 9

Convert the number by repeatedly squaring until we get the unit digit as 1, and then applying the trick of finding the last two digits of number with unit digit 1 as explained above.

Example: *Find the last two digits of *\({79^{64}}\)

Solution:

Observe that the last two digits of \( {79^2} \equiv {21^2} \equiv 41\)

Therefore, the last two digits of \({79^{64}} \equiv \) last two digits of \({\left( {{{79}^2}} \right)^{32}} \equiv \) \({41^{32}} \equiv 81\)

Example: Find the tens place digit of \({87^{73}}\)

Solution:

The last two digits of \({87^{73}}\)

\( \equiv \;last\;two\;digits\;of\;87 \times {\left( {{{87}^2}} \right)^{36}}\) \( \equiv 87 \times {\left( {69} \right)^{36}} \equiv 87 \times {\left( {{{69}^2}} \right)^{18}}\)

\( \equiv 87 \times \left( {{{61}^{18}}} \right) \equiv 87 \times 81 \equiv 47\)

Hence the tens place digit is 4.

### Special Case of finding the last two digits of number raise to the power of 7:

Question: What are the last two digits of \(7^{2008}\)? [CAT 2008]

a. 21

b. 61

c. 01

d. 41

e. 81

**Solution:**

### Finding the last two digits of even number

Things to note is that the last two digits of

- \({24^{Odd}}\) is always 24.
- \({24^{Even}}\) is always 76
- \({76^{any\;natural\;number}}\) is always 76.

Now we will take an example to use the above observation.

Example: Find the last two digits of \({2^{34}}\)

Solution:

\({2^{34}} \equiv {\left( {{2^{30}}} \right)^4}\)

\( \equiv {\left( {{2^{10}}} \right)^3} \times {2^4}\)

\( \equiv {24^3} \times {2^4}\)

\( \equiv 24 \times 16\)

\( \equiv 84\)

Example: Find the last two digits of \({62^{48}}\)

Solution:

Observe that \({62^{48}}\) can be written as \({2^{48}} \times {31^{48}}\)

Or last two digits of \({62^{48}}\) = last two digits of \({2^{48}}\) \( \times \) last two digits of \({31^{48}}\)

Last two digits of \({2^{48}} \equiv {\left( {{2^{10}}} \right)^4} \times {2^8}\)

\( \equiv {24^4} \times {2^8}\)

\( \equiv 76 \times 56 \equiv 56\)

And

Last two digits of \({31^{48}} \equiv 41\)

Therefore, the last two digits of \({62^{48}} = 56 \times 41 = 96\).

### Finding the tens place digit of a number ending with 0 or 5

Any number with its units digit as 0 when raised to any power has 00 as its last two digits.

Below table shows rules to find last two digits of numbers ending in 5

Tens digit of number | Units digit of power | Last two digits | Example |

Even | Even | 25 | 25 = 6^{2}25 |

Even | Odd | 25 | 25 = 156^{3}25 |

Odd | Even | 25 | 15 = 2^{2}25 |

Odd | Odd | 75 | 15 = 33^{3}75 |

Example: Find the last two digits of \({135^{123}}\)

Solution:

Tens digit of given number (135) = odd

The units digit of power (123) = odd

From the above table, we can see that for odd-odd combination last two digits will be **75. **

### Articles on Number System

- Number system syllabus and preparation tips for CAT exam
- HCF and LCM Concepts
- Number of Factors of a Number
- Concepts of Remainders
- Remainders - Fermat Theorem
- Chinese Remainder Theorem
- 3 Steps to Find Last Two Digit Numbers
- Factorials - Number of Trailing Zeros
- Base System Concepts and Questions

## 27 Responses

3 power 600 answer

01

Thanks bhai

I don’t quite understand the concept of bionomial expansion you mentioned for verification.

Good

01

3^600=((50-47)^2)^300=(2500+4700-2209)^300=(09)^300=((50-41)^2)^150=(2500+4100-1681)^150=(81)^150=00 (ten’s place of 81(i.e 8) * unit place of power(i.e 0))

so the two digit number turns out to be unit place of 00 and unit place of base i.e 01

3^600= (3^4)^150

=(81)^150

=01

01

Marvellous

how can we find 99 power 99

Last two digits will be 99

Since 99^99=9^99×11^99

=(9^2)^49×9×91[applying 2 digit numbers endind with 1]

=(81)^49×19

=21×19

=99

(18763)KA POWER 237 H FIND LAST TWO DIGIT

37

99^99 can be written as 99^98 * 99

since we can write 99 as (100-1)

therefore, ((100-1)^2)^49 * 99

= (01)^49 * 99 [applied two digit rule]

= 01 * 99

= 99

Last two digit of 99 ^99 is 99

99^99

= (100-1)^99

= 100^99 – 99C1 . 100^98 + 99C2. 100^97 – ……………….. – 99C99. 100^0. (1)^99

= ………………0000000 – 1

= ………………9999999999

Therefore, last two digits of 99^99 = 99

(18763)KA POWER 237 H FIND LAST TWO DIGIT

83

Solution?

I am getting 23

83

@Laduu 83

my last 2 digit is 65

If possible Pls help what is my full number Pls it’s too urgent

hi

Wait so what is 2017 to the power of 2017 ?

77

150^134 ,

Last 2 non zero digits ?

please answer (7^14)14