Latest TCS Aptitude Question SOLUTION: (72)^21/ 25

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(#M40039415) TCS Numerical Ability question Keep an EYE

(72)^21/ 25

Asked In TCS PARUL JOSHI (10 years ago)

Unsolved Read Solution (11)

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Read Solution (11) :

khushboo  (10 years ago)

using -ve remainder theorm
(75-3)^21/25
(-3)^21/25=(-2)^7/25=-3/25=22
so ans is 22

Is this solution Helpfull? Yes (12) | No (18)  

GOGULA VENKATA KRISHNA REDDY  (10 years ago)

last 2 digits of 72^21 is 72.
so xxx72/25=22
so ans is 22.
Explanation to find last 2 digits of 72^21 is
72=(2^3)*(3^2) --> 72^21=(2^63)*(3^42) now find last 2 digits of 2^63 and 3^42

2^63=(2^60)*(2^3)=((2^10)^6)*(2^3)
last 2 digits of 2^10 is 24^6 6 is even so 76 is last 2 digits of 2^60
76*8=08 is last 2 digits

3^42=((3^4)^10)*9=(81^10)*9
last 2 digits of 81^10 is 1
1*9=9
so now last 2 digits of 2^63*3^42 is 8*9=72

so finally 72%25=22

Is this solution Helpfull? Yes (2) | No (3)  

khushboo  (10 years ago)

using -ve remainder theorm
(75-3)^21/25
(-3)^21/25=(-27)^7/25=(-25-2)^7/25=(-2)^7/25
=-128/25=-3/25=22
so ans is 22

Is this solution Helpfull? Yes (2) | No (5)  

saurabh kumar  (10 years ago)

ans is 15.

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GOGULA VENKATA KRISHNA REDDY  (10 years ago)

can u explain the -ve remainder theorem clearly khushboo

Is this solution Helpfull? Yes | No (2)  

mounika  (10 years ago)

22.i think

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virendra tiwari  (10 years ago)

rem 22,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

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Sourav ganguly  (10 years ago)

applying remainder theorem
12^21/25
12^3^7/25
12*12*12)^7/25
19*12)^7/25
228^7/25
3^7/25
3^3*3^3*3/25
2*2*3/25
so remainder is 12

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aman sharma  (10 years ago)

(75-3)^21/25
=remainder -3
or we can say 22

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Kethineni Vinodh  (10 years ago)

using -ve remainder theorm
(75-3)^21/25
(-3)^21/25=(-27)^7/25=(-25-2)^7/25=(-2)^7/25
=-128/25=-3/25=22
so ans is 22

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purva  (10 years ago)

22 is right answer
Khushboo has done correct using -ve remainder theorem

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