# Number System Tutorial Part III: Factors, Multiples, Unit Digit and Last Two Digits of Exponents

## How to find last two digits of a^b

If 'a' and 'b' are two natural numbers, then it is possible to find out the last two digits of the expression a^b. We can classify the situations as per the unit digit of the base value 'a'.

### Last two digits of exponents of a number which ends in '0'

Example: 20^n
If n = 1 -> last two digits are the last two digits of the base itself, i.e. 20
If n > 1 -> last two digits are always '00'.

### Last two digits of exponents of a number which ends in '1'

• Unit digit of a^b is always 1.
• Tens digit of a^b = "unit digit of"[ "tens digit of 'a'" xx "unit digit of 'b'"]
Find the last two digits of 21^34?
Unit digit = 1
Tens digit = 2 xx 4 = 8
Therefore the last two digits of 21^34 are 81.
Find the last two digits of 1341^73?
Unit digit = 1
Tens digit = 4 xx 3 = 12 -> 2,(unit digit of 12)
Therefore last two digits of 1341^73 = 21

### Last two digits of exponents of a number which ends in '5'

• If the exponent is 1, then the last two digits are the last two digits of the given number itself.
• If the exponent is greater than 1, then the last two digits are always '25'.
Find the last two digits of 1375^46
1375 ends in 5 and the exponent is greater than 1, therefore the last two digits of the given expression are 25.

### Last two digits of exponents of a number which ends in 3, 7 or 9

• 3^4, 7^4 and 9^2 are ending in 1.
• Convert the base in the form of ending with 1, through the conversion of the exponent.
• After changing the unit digit of the base to 1, then apply the method which we discussed for numbers ending in 1.

Let's look at an example to see how to find last two digits of an exponent of a number which ends in 7

Find the last two digits of 17^244?
First convert the base in such a way that it ends with 1.
17^244 = (17^4)^61
= (17^2 xx 17^2)^61
= (289 xx 289)^61
= ("___21")^61; Now base ends in 1, we can apply the procedure to find last two digits of exponents of a number which ends in 1
Unit digit = 1
Tens digit = 2 xx 1 = 2
:. "Last two digits of " 17^244 = 21

For finding the last two digits of the exponents of a number which ends in 3, 7 or 9, the following patterns will help you to make a quick approach.

#### Pattern I: Last two digits of 4th power of the numbers which end in 3.

03^4 -> 81 \leftarrow 53^4
13^4 -> 61 \leftarrow 63^4
23^4 -> 41 \leftarrow 73^4
33^4 -> 21 \leftarrow 83^4
43^4 -> 01 \leftarrow 93^4
Let's look at an example to see how to apply these patterns in practise.

Find the last two digits of 133^287 ?
"Last two digits of " 133^287 -> "Last two digits of " (33^4)^71 xx 33^3
From the above pattern it is easy to catch the last two digits of 33^4, and those are 21.
-> "Last two digits of " 21^71 xx 33^3
-> "Last two digits of " 21 xx 37 (because the last two digits of 21^71 are 21 and that for 33^3 = 33 xx 33 xx 33 -> 37)
:. "Last two digits of "133^287 " are " 77.

#### Pattern II: Last two digits of 4th power of the numbers which end in 7.

07^4 -> 01 \leftarrow 57^4
17^4 -> 21 \leftarrow 67^4
27^4 -> 41 \leftarrow 77^4
37^4 -> 61 \leftarrow 87^4
47^4 -> 81 \leftarrow 97^4

Find the last two digits of 157^45 ?
"Last two digits of " 157^45 -> "Last two digits of " (57^4)^11 xx 57
"Last two digits of " (01)^11 xx 57 (From the above pattern we can see that last two digits of 57^4 is 01)
= 57
Find the last two digits of 2067^158 ?
"Last two digits of " 2067^158 -> "Last two digits of " (67^4)^39 xx 67^2
-> "Last two digits of " 21^39 xx 89
-> "Last two digits of " 81 xx 89 = 09

#### Pattern III: Last two digits of square of the bases which ends in 9.

09^2 -> 81 \leftarrow 59^2
19^2 -> 61 \leftarrow 69^2
29^2 -> 41 \leftarrow 79^2
39^2 -> 21 \leftarrow 89^2
49^2 -> 01 \leftarrow 99^2

Find the last two digits of 39^121?
39^121 -> (39^2)^60 xx 39
->" Last two digits of " (21)^60 xx 39 (From the above pattern we can see that last two digits of 39^2 are 21)
->" Last two digits of " 01 xx 39
Therefore the last two digits of 39^121 are 39.

### Last two digits of exponents of a number which ends in 2, 4, 6 and 8

• Convert the given bases in the form of 2^10
• (2^10)^"even" will end in 76.
• (2^10)^"odd" will end in 24.
Find the last two digits of 2^70?
2^70 = (2^10)^7 = (2^10)^"odd"
Therefore the last two digits are 24.
Find the last two digits of 8^25?
8^25 = (2^3)^25 = 2^75
= (2^10)^7 xx 2^5
-> " Last two digits of " 24 xx 32 = 68
Find the last two digits of 108^24?
108^24 = (4 xx 27)^24
= (2^2)^24 xx 27^24
= 2^48 xx 27^24
=(2^10)^4 xx 2^8 xx 27^24
Last two digits of (2^10)^4 = 76
Last two digits of 2^8 = 56
Last two digits of 27^24 = (27^4)^6 -> " Last two digits of " (41)^6 -> 41
Therefore the last two digits of 108^24 = Last two digits of [76 xx 56 xx 41] = 96

### Tips for quick calculation

• 76^n is always end in 76 for any natural number values for n.
• 24^"even" will end in 76.
• 24^"odd" will end in 24.
• 76 xx 2^n will end in the last two digits of 2^n, where n >= 2.
Example :
76 xx 32 ends in 32.
76 xx 1024 ends in 24.
76 xx 16 ends in 16.
• 76 xx " any multiple of 4" will end in the last two digits of the multiple of 4.
Example:
76 xx 04 ends in 04.
76 xx 88 ends in 88.
• 76 xx " any even number but not a multiple of 4" will end in 50 + "last two digits of the number"
Example:
76 xx 2 ends in 50 + 2 = 52
76 xx 6 ends in 50 + 6 = 56
76 xx 22 ends in 50 + 22 = 72
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