Latest Elitmus Aptitude Question SOLUTION: What are last two digits of 7^2008 A. 21 B. 61 C. 01 D. 41

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(#M40037979) ELITMUS Category question Keep an EYE

What are last two digits of 7^2008
A.
21
B.
61
C.
01
D.
41

Asked In Elitmus Liti (9 years ago)

Unsolved Read Solution (9)

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Read Solution (9) :

Prateek Pandey  (9 years ago)

C.01

7^1 last two digits will be 07

7^2 last two digits will be 49

7^3 last two digits will be 43

7^4 last two digits will be 01

this pattern will be reeating after every 4 terms i.e

7^5 last two terms will be 07 again & so on

hence for 7^2008
since the power is completely cancelled out from 4 it will be similar to that of

7^4 i.e. 01

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DIMPLE SINGH  (9 years ago)

C. 01

7^2008 = (7^4)^502
= 2401^502
=...01^502
= ....01

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Sanjay Patel  (9 years ago)

01
2008,last two digit is a multiple of 4.so whatever last two digits of 7*7*7*7=2401,it will become the last two digits of the question.so C is final answer.

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sumit mondal  (9 years ago)

((7)^502)^4)....

means

7^1=7
7^2=49
7^3=343
7^4=2401.....

07,49,43,01.....are repeated a 7 power

hence correct answer is 01....(c)

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Pooja bala singh  (9 years ago)

B.21

(7^3)^669*7=3*7=21

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jalpan  (9 years ago)

@Pooja How can u explain ??

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Akash Poddar  (9 years ago)

[0*8][7^8]
01

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Abhishek  (9 years ago)

i think its 01 ...can anyone please explain the answer ??

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Kriti Jaiswal  (9 years ago)

C.01 divide the no. by 100 and find the remainder

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