Latest TCS Aptitude Question SOLUTION: Find last digit.(1023^3923)+(3087^3927)

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Find last digit.(1023^3923)+(3087^3927)

Asked In TCS nehaghosh (10 years ago)

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Sourangsu Bandyopadhyay  (10 years ago)

The solution can derived as:
3923 % 4=3
3927 % 4=3 (also)

Now last digit of 1023 is 3, so 3^3923->3^(3923 % 4)->3^3=27(take last digit i.e 7)
Similarly,last digit of 3087 is 7, so 7^3927->7^(3923 % 4)->7^3=343(take last digit i.e 3)

Add 7+3= 10
(Take last digit of 10 i.e. 0)

Thus, the answer is 0.

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RAKESH  (10 years ago)

last digit of the exp will be same as the last digit of 3^3923 + 7^3927
3923=4*980+3 & 3927=4*981+3 so last digit of 3^3923 + 7^3927=last digit of 3^3 + 7^3
i.e 27+343
hence unit digit will be 7+3=10
ans is zero

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sriparna chakraborty  (10 years ago)

3^3+7^7
=7+3
=10
so last digit of this equation is 0

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juhe rawat  (10 years ago)

ans: 0
3^3=81 2*3=6
7^7=49 8*7=56

681+5649=last digit comes to 0

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shiwani priya  (10 years ago)

3^3923+7^3927=7+3=0

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ronak basaiwal  (10 years ago)

1023^3923= 7 last digit
3087^3927= 3 last digit
therefore,last digit is 0.

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Tarun Sharma  (10 years ago)

last digit will be :: 0

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suparna das  (10 years ago)

last digit is 0

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Guru Pranav  (10 years ago)

0 i s the answer !!!

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