Latest TCS Aptitude Question SOLUTION: 4. Find last digits of : (1023^3923)+ (3087^3927)

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4. Find last digits of : (1023^3923)+ (3087^3927)

Asked In TCS Bishal Pandey (10 years ago)

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Read Solution (6) :

Nirupam Das  (10 years ago)

Any number with last digit 3 and 7 has cycle of 4. It means after evry fourth power the last digits repeat. For 3 its 3,9,7,1,3,9.... and for 7 its 7,9,3,1,7,9... And 1^anything is 1. So it can be written as
3^(4n+3)+7^(4n+3). 3^4n and 7^4n will result in last digit being 1, so omitting it we are only left with 3^3+7^3 whose last digits are 7+3=10 and hence anser is 0.

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prabhat kumar  (10 years ago)

(1023)^1023/4=(1023)^3
last digit of base is 3,so 3^3=27 last digit is 7
and another (3087)^3927/4=(3087)^3
last digit is 0
last digit of this number base is 7,so 7^3=343
last digit is 3
so,last digit of given problem is (7+3=10) 0.

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Anand  (10 years ago)

3^3+7^3=7+3=10
SO LAST DIGIT IS 0(ANS)

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Sri Reddy  (10 years ago)

In 1023 the last digit is 3 which has cycle of 4..so divide 3923 with 4 the remainder is 3..3^3=27 last digit is 7
In 3087 the last digit is 7 which has cycle of 4 so divide 3927 with 4 remainder is 3....7^3=343 last digit is 3
finally 7+3=10 soo last digit is 0

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Anirban banerjee  (10 years ago)

3923%4=3 and here the base digit is 3 so 3^3=27
similarly 3927%4=3 and 7^3=343
now add dos values 343+27=370
so the last digit is 0(Ans)

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ARUN GUPTA  (10 years ago)

3^3+7^3=7+3=10
SO LAST DIGIT IS = 0

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