a.Last two digits of numbers which end in
one.

b. Last two digits of numbers which end in 3, 7
and 9

c. Last two digits of numbers which end in
2

d. Last two digits of numbers which end in 4, 6
and 8

Before we start, let me mention binomial theorem in
brief as we will need it for our calculations.

Last two digits of numbers ending in 1

Letâ€™s start with an
example.

**What are the last two digits
of 31**^{786}?

Solution: 31^{786} = (30 +
1)^{786} =
^{786}C_{0}
x 1^{786} +
^{786}C_{1
} x
1^{785}
x (30) +
^{786}C_{2}
x 1^{784} x 30^{2} +
…, Note that all the terms after the
second term will end in two or more zeroes. The first two terms are
^{786}C_{0
} x 1^{786} and
^{786}C_{1
} x
1^{785}
x (30). Now, the
second term will end with one zero and the tens digit of the second
term will be the product of 786 and 3 i.e. 8. Therefore, the last
two digits of the second term will be 80. The last digit of the
first term is 1. So the last two digits of 31^{786} are 81.

Now, here is the shortcut:

Here are some more examples:

Find the last two digits of **41**^{2789}

In no time at all you can calculate the answer to be
61 (4 × 9 = 36. Therefore, 6 will be the tens digit and one will be
the units digit)

Find the last two digits of **71**^{56747}

Last two digits will be 91 (7 × 7 gives 9 and 1 as
units digit)

Now try to get the answer to this question within 10
s:

Find the last two digits of **51**^{456} x
**61**^{567}

The last two digits of 51^{456} will be 01 and the
last two digits of 61^{567} will be 21.
Therefore, the last two digits of 51^{456}
x 61^{567} will be the last two
digits of 01 × 21 = 21

Last two digits of numbers ending in
**3, 7 or 9**

Find the last two digits of **19**^{266}.

19^{266} =
(19^{2})^{133}.
Now, 19^{2} ends
in 61 (19^{2} = 361)
therefore, we need to find the last two digits of
(61)^{133}.

Once the number is ending in 1 we can straight away
get the last two digits with the help of the previous method. The
last two digits are 81 (6 x 3 = 18, so the tens digit will be 8
and last digit will be 1)

Find the last two digits of **33**^{288}.

33^{288} =
(33^{4})^{72}.
Now 33^{4} ends in 21
(33^{4} =
33^{2} x
33^{2} = 1089
x 1089 = xxxxx21)
therefore, we need to find the last two digits of 21^{72}. By the previous
method, the last two digits of 21^{72} = 41 (tens digit = 2 ×
2 = 4, unit digit = 1)

So hereâ€™s the rule for finding the last two digits
of numbers ending in 3, 7 and 9:

Now try the method with a number ending in 7:

Find the last two digits of **87**^{474}.

87^{474} =
87^{472} x
87^{2} =
(87^{4})^{118}
x 87^{2} = (69 x 69)^{118} x 69 (The last two digits of
87^{2} are 69) =
61^{118} x
69 = 81 x
69 = 89

If you understood the method then try your hands on
these questions:

Find the last two digits of:

1.
27^{456}

2. 79^{83}

3. 583^{512}

Last two digits of numbers ending in 2, 4, 6 or 8

There is only one even two-digit number which always
ends in itself (last two digits) - 76 i.e. 76 raised to any power
gives the last two digits as 76. Therefore, our purpose is to get
76 as last two digits for even numbers. We know that 24^{2} ends in 76 and
2^{10} ends in 24.
Also, 24 raised to an even power always ends with 76 and 24 raised
to an odd power always ends with 24. Therefore, 24^{34} will end in 76 and
24^{53} will end
in 24.

Letâ€™s apply this
funda:

**Find the last two digits
of** **2**^{543}.

2^{543} =
(2^{10})^{54}
x 2^{3} =
(24)^{54} (24 raised to an even
power) x 2^{3} = 76 x 8 = 08

(**NOTE:** Here if you need to multiply 76 with
2^{n}, then you can straightaway write the
last two digits of 2^{n} because when 76
is multiplied with 2^{n} the last two
digits remain the same as the last two digits of
2^{n}. Therefore, the last two digits of
76 x 2^{7} will be the last two digits of
2^{7} = 28)

Same method we can use for any number which is of the
form 2^{n}. Here
is an example:

Find the last two digits of **64**^{236}.

64^{236} =
(2^{6})^{236} =
2^{1416} =
(2^{10})^{141}
x 2^{6} =
24^{141} (24 raised to odd power)
x 64 = 24
x 64 = 36

Now those numbers which are not in the form of 2n can
be broken down into the form 2n Â´ odd number. We can find the last
two digits of both the parts separately.

Here are some examples:

**Find the last two digits of
62**^{586}.

62^{586} =
(2 x 31)^{586} =
2^{586} x
3^{586} =
(2^{10})^{58}
x 2^{6} x 31^{586} =
76 x 64
x 81 = 84

**Find the last two digits of
54**^{380}.

54^{380} =
(2 x 3^{3})^{380}
= 2^{380} x
3^{1140} =
(2^{10})^{38}
x (3^{4})^{285}
= 76 x 81^{285} = 76 x 01 = 76.

**Find the last two digits of
56**^{283}.

56^{283} =
(2^{3} x
7)^{283} =
2^{849} x
7^{283} =
(2^{10})^{84}
x 2^{9} x (7^{4})^{70}
x 7^{3} = 76 x 12 x
(01)^{70}
x 43 = 16

**Find the last two digits of
78**^{379}.

78^{379} =
(2 x 39)^{379} =
2^{379} x
39^{379} =
(2^{10})^{37}
x 2^{9} x (39^{2})^{189}
x 39 = 24
x 12 x 81 x
39 = 92

## Post Comments

Rohityadav876– Mon, 06 Oct 2014 10:38:25 -0000(347)^347 => (347^4)^86*(347)^3

=> (xxxx81)^86*(347)^3 => (8*6=48 so 8 will be our tens) for one we have to do some more like this(347)^3=(xxxx24)*(81)= so our last digit will be clearly 4 this will be our one

so last two digit wil be 84

Rohityadav876– Mon, 06 Oct 2014 10:41:16 -0000(347)^347 => (347^4)^86*(347)^3

=> (xxxx81)^86*(347)^3 => (8*6=48 so 8 will be our tens) for one we have to do some more like this(347)^3=(xxxx24)*(81)= so our last digit will be clearly 4 this will be our one

so last two digit wil be 84

Rohityadav876– Mon, 06 Oct 2014 10:41:42 -0000(347)^347 => (347^4)^86*(347)^3

=> (xxxx81)^86*(347)^3 => (8*6=48 so 8 will be our tens) for one we have to do some more like this(347)^3=(xxxx24)*(81)= so our last digit will be clearly 4 this will be our one

so last two digit wil be 84