Latest TCS Aptitude Question SOLUTION: What are the last two digits of expression 5306^214?

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What are the last two digits of expression 5306^214?

Asked In TCS Nakshatra Mukhopadhyay (8 years ago)

Unsolved Read Solution (9)

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leelakrishna  (8 years ago)

96
hence, 6^214=6^200*6^14
=(6^2)^100*(6^2)^6*6^2
=>>6*6*36
as the expansion of (6^2) must consist 36
so last 2 digit =>>96(answer)

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s.h.h.praveen  (8 years ago)

24
multiply last 2 no.s 6*4=24

Is this solution Helpfull? Yes (3) | No (9)  

milan jena  (8 years ago)

5306=2*7*379
So (5306)^214=(2*7*379)^214
2^214=(2^10)^21 *2^4 = (1024)^21 * 16 = 24*16=384
7^214=(7^4)^53 * 7^2 =(49*49)^53 * 7^2
=(2401)^53 * 49 = 01*49=49
(379)^214= (79)^214 = 21^214 =81

84*49*81=96

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AKASH SHIV SHUKLA  (8 years ago)

06 would be the answer

Is this solution Helpfull? Yes (1) | No (6)  

sameer sin  (8 years ago)

take the lost two digits and raise to power
41^3=68921
61^1=61
add both lost digit will be 82.

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adi  (8 years ago)

96 would be the ans

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sugandha sinha  (8 years ago)

ans will be: 06
0*4= 0
6^4 = 6

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SHUVRAJYOTI DEB ROY  (8 years ago)

CORRECT ANSWER IS 96
CYCLICK ORDER OF 6 IS 5
SO 214/5, REM = 4
ie 6^4 = 96

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  (6 years ago)

76 power any number is 76 and multiplication of 76 with any 6 multiples ending with 6 gives same last two digits.
eg:
76^1 = 76
76^2 = 5776
76^3 = 438976 ....... last two digits are 76 only
and
76*36 = 2736 ,76*96 = 7296....
6^214 = 6^(5*42)*6^4 = 76*96 = 96 (since 6^5 gives 76)
last two digits are 96.

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