Last two digit of the number (129)^129 will be

SANTOSH SAHOO   9 years AGO

(-1+130)^129
=prove using binomial theorem
=-1+130*129
=167769
last 2 digit is 69

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Yathiraj   9 years AGO

There are 10 possible last 2 digits with powers of 129
power = 1, last 2 digits = 29
power = 2, last 2 digits = 41
power = 3, last 2 digits = 89
power = 4, last 2 digits = 81
power = 5, last 2 digits = 49
power = 6, last 2 digits = 21
power = 7, last 2 digits = 09
power = 8, last 2 digits = 61
power = 9, last 2 digits = 69
power = 10, last 2 digits = 01

since the power is 129, 129 mod 10 = 9
and from the above list, last 2 digits will be 69.

Option 7

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Meenakshi   9 years AGO

(129^2)^64 * 129
(16641)^64 * 129
(41)^64 * 129
last two digit of (41)^64 is 61 (multiply tens digit of a number with last digit of the exponent to get the tens digit 4*4=16.unit digit is equal to one)
so,61*129
=>7869
so ans is 69

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SHAHEEN   9 years AGO

in powers of 129 we get last digits as 69
so ans is 69---option 7.

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KUMAR HARSH   9 years AGO

(129^2)^64*129
the last two digit of (129^2)^64=61
hance last two digit of (129)^129=last two digit of 61*129 =69

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siddharth prakash singh   9 years AGO

find remainder of (129^129)/100;
=>29^129/100
use cyclicity now...
we get answer as 69.

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KALYAN KUMAR REDDY VALLURU   9 years AGO

as 129/4 will give the reminder as 1
so units digit must be 9

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aman   9 years AGO

129^129

29 ^129
(29^(2*64)+1) = 29^(2*64) . 29

29^2 last 2 digit 41
41^64 . 29
41 ^ 64 last two digit 61
61*29 = last two digit 69

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CHAYAN KUMAR   9 years AGO

because last two digit of power 1,2,3,4,5,6,7,8,9,10,11 are respectively 29,41,89,81,49,21,09,61,69,01,29
so we can see that last two digit is repeated after (121)^11
so (129)^121 last two digit will be 29 & (129)^8 last two digit will be 61
so (121)^129 last two digit will be 61
so option 7 is the right one

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kumar chayan   9 years AGO

last two digit of (129)^11=29 so repeated after every every 11 power so 129^121=29 & 129^8=61 (last two digit) so (129)^129=61 which is required ans

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sompon temtheerakij   9 years AGO

the last two digit is
29,41,89,81,49,21,09,61,69,01,29,41.......
129th is 69

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vibhu raj gupta   9 years AGO

(129^2)^64*129
(16641)^64*129
last two digits of 16641 is mul tens digit with unit digit of power 4*4=16
take 6 and for unit place 1 is i.e. 61
now 61*129
last two digit is 69

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Settipalli.Anil Reddy   9 years AGO

129^129
The last two digits are 69

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Thirupathi Reddy   9 years AGO

129^129
The last two digits are 69

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ARUP GHOSH HAZRA   9 years AGO

129^129
= ((29)^2)^64 *29
=41^64 *29
the above expression can be written as ;
=(41)^60 *(41)^4 *29
=01 *61 *29
=69

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J.DHANDAPANI   9 years AGO

By Binomial theorem,
(a+x)^n= nc0 a^n x^ 0 + nc1 a^(n-1) x ^1 + …x100 + …x1000+…
(-1 + 130)^129 = 1x (-1)^129 x130^0 +129 x(-1)^128 x 130 + .....
= -1 + 16770 +.....
=.......16769

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Abhijeet Digarse   9 years AGO

(129)^129 => ((29)^2)^64 × (29)^1
=> (41)^64 ×29
=> 61 × 29
=> 69. ( here we going only with last two digits after every step)

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shareef   9 years AGO

29^9 = .... 69

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Math Lovers   9 years AGO

(129)^129 =(130-1)^129
= ......... +130*129 - (1)^129 = 130*129 - (1)^129
= -------70 - 1= --------------69

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aman aditya   9 years AGO

number last two digit
29^2 41
29 ^4 81
29^8 61
29^16 21
29^32 41
29^64 81
29^128 61

now
61*29 69

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Bheemaray s vangi   9 years AGO

29^1=29-> 29
29^2=841-> 41
29^4=41^2-> 81
29^8=81^2-> 61
29^16=61^2-> 21
29^32=21^3-> 41
29^64=81
29^128=61
61*29=1769
last two digits are 69

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shabana banu shaik   9 years AGO

29^1 = 29 --> 29
29^2 = 841 --> 41
29^4 = 41^2 = 1681 --> 81
29^8 = 81^2 = 6561 --> 61
29^16 = 61^2 = 3721 --> 21
29^32 = 21^2 = 441 --> 41 * at this point the pattern will just repeat:
29^64 --> 81
29^128 --> 61

Now split 129 into powers of 2.
129 = 128 + 1

So you want the results for 29^128 * 29^1

But again just use the last two digits
61 x 29
= 1769 (mod 100) = 69

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vijay   9 years AGO

129^129
(130 - 1)^129
In bionomial expansion
(a + b)^n
(r +1)th term is given by
t(r + 1) = nCr * a^(n-r) * b^r
and has (n + 1)th term

And I think that last two digits can be obtained by last two terms so lets see
here
a = 130, b = -1, n = 129
t(129) = 129C128 * 130^(129 - 128) * (-1)^128
t(129) = 129 * 130 * 1 = 16770

t(130) = 129C129 * 130^(129 - 129) * (-1)^129
t(130) = 1 * 1 * -1 = -1

t(129) + t(130) = 16770 - 1 = 16769
So
last two digits are
69

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Surit Datta   9 years AGO

129^129 (mod 100)

Because any digit beyond the smallest two won't affect the tens or ones place, we can think of this equivalently as:
29^129 (mod 100)

In general, the efficient method for computing a modular exponent like the one above is to start by repeatedly squaring, and reducing after each square, recording only the last two digits, like so:

29^1 = 29 --> 29
29^2 = 841 --> 41
29^4 = 41^2 = 1681 --> 81
29^8 = 81^2 = 6561 --> 61
29^16 = 61^2 = 3721 --> 21
29^32 = 21^2 = 441 --> 41 * at this point the pattern will just repeat:
29^64 --> 81
29^128 --> 61

Now split 129 into powers of 2.
129 = 128 + 1

So you want the results for 29^128 * 29^1

But again just use the last two digits
61 x 29
= 1769 (mod 100) = 69

so option 7
that's it.

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Divya jain   9 years AGO

For last two digit we can write it as
(29)^129 = ((29)^2)^64 * 29
=((41)^2)^32 * 29
= ((81)^2)^16 * 29
= ((61)^2)^8 * 29
= ((21)^2)^4 * 29
= ((41)^2)^2 * 29
= ((81)^2) * 29
= 61 * 29
last two digit will be = 69

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Supriya Mondal   9 years AGO

(129)^129=(129^2)^64*129
=(16641)^64*129
We know that from the formula that last two digit of (16641)^64 will be 61
So, last two digit of the above product will be 69
So, option 7)69 is the right answer

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ABINAYA R   9 years AGO

29^1 = 29 --> 29
29^2 = 841 --> 41
29^4 = 41^2 = 1681 --> 81
29^8 = 81^2 = 6561 --> 61
29^16 = 61^2 = 3721 --> 21
29^32 = 21^2 = 441 --> 41 * at this point the pattern will just repeat:
29^64 --> 81
29^128 --> 61

Now split 129 into powers of 2.
129 = 128 + 1

So the results for 29^128 * 29^1

But again just use the last two digits
61 x 29
= 1769 (mod 100) = 69

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RAKESH   9 years AGO

(129)^129
for last two digit(29)^129
(30-1)^129 =--------------+ 129c128*30 - 1 = 70-1 = 69

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sravani   9 years AGO

It would be 69...

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Jeevitha   9 years AGO

129^129=129*129^128
the last two digits of 129^128 is 61
and by multiplying 61 with 129 we will get the last two digits as 69

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