Latest m4math Aptitude Question SOLUTION: Last two digit of the number (129)^129 will be

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(#M40130850) M4MATH Numerical Ability question last two digit Keep an EYE

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

Asked In m4math MRITUNJAY (9 years ago)

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Read Solution (6) :

Surit Datta  (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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Jay Kumar  (9 years ago)

(129)^129 = 9^129 = (9)^1*(9)^128 = 9*1 = 9 Answer

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Jay parmar  (9 years ago)

(129)^129
= 9^129
= (9)^1*(9)^128
=(-1+10)^128 * 9
=128c0* (-1)^128 *10c0 + 128c1 * (-1)^127 * 10c1
= 1-1280 = -1279
Now
= -1279*9
=11511
means last two digit will be 11
option 2 is correct

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poojitha  (9 years ago)

can u pls explain @jay kumar

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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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shubham garg  (8 years ago)

(129)^129
= (100+29)^129/100 = (29)^129
now, we can also write it as:
(100-71)^129/100
(-71)^129 = -39
therefore, 100-39 = 61
thus last two digit number is 61

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