Latest TCS Aptitude Question SOLUTION: Find last digit.(1023^3923)+(3087^3927) (what is d tricks of finding these last digit sums??..plzz explain)

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Find last digit.(1023^3923)+(3087^3927) (what is d tricks of finding these last digit sums??..plzz explain)

Asked In TCS TCS 2014 (10 years ago)

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Sayan Banerjee  (10 years ago)

last digit:
Cycle of 3:
1) 3^1= 3 is the last digit.
2) 3^2= 9 is the last digit.
3) 3^3= 7 is the last digit.
4) 3^4= 1 is the last digit.
an then for 3^5 onward's the cycle repeats.
Therefore, for 1023^3923 => 3^3923 = 3^(4*980 +3) So remainder is 3 when the exponent 3923 is divided by 4. So from point '4)' we move 3 steps in cyclic order to go to point '3)'. So last digit of the first part is 7.
Similarly cycle for 7 for the 2nd part.
1) 7^1= 7 is the last digit.
2) 7^2= 9 is the last digit.
3) 7^3= 3 is the last digit.
4) 7^4= 1 is the last digit.
an then for 7^5 onward's the cycle repeats.
And hence find the last digit in the similar fashion. It comes to 3.
So total is 7+3= 10. So last digit is 0.

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TCS 2014  (10 years ago)

1023^3923(3923=4n+3) so last digit 7
3087^3927(3927=4n+3)so last digit 3
now adding we get last digit 0

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saurabh agarwal  (10 years ago)

last digit:
Cycle of 3:
1) 3^1= 3 is the last digit.
2) 3^2= 9 is the last digit.
3) 3^3= 7 is the last digit.
4) 3^4= 1 is the last digit.
an then for 3^5 onward's the cycle repeats.
so cyclic for 3 is 4.
Therefore, for 1023^3923 => 3^3923 = 3^(4*980 +3),3^3=27 So last digit of the first part is 7.
Similarly cycle for 7 for the 2nd part.
1) 7^1= 7 is the last digit.
2) 7^2= 9 is the last digit.
3) 7^3= 3 is the last digit.
4) 7^4= 1 is the last digit.
so cyclic of 7 is 4.
And hence find the last digit in the similar fashion. It comes to 3.
So total is 7+3= 10. So last digit is 0.

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moumita ghose  (10 years ago)

1st part last digit is 7 as 3^3. 2nd part: 7^3 as(4 is universil cyclicty. so 7 didided by 4= 3. and 7^3=lst digit 3.). so 7+3=0 so ans 0 is d last digit.

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erupuru poojitha  (10 years ago)

last digit is 0

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nikhita krishna  (10 years ago)

lst digit of frst term=7, lst digit of 2nd term 3, 7+3=10...overall lst digit=0

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Aditya Kushwaha  (10 years ago)

LAST DIGIT:
Cycle of 3:
1) 3^1= 3 is the last digit.
2) 3^2= 9 is the last digit.
3) 3^3= 7 is the last digit.
4) 3^4= 1 is the last digit.
an then for 3^5 onward's the cycle repeats.
Therefore, for 1023^3923 => 3^3923 = 3^(4*980 +3) So remainder is 3 when the exponent 3923 is divided by 4. So from point '4)' we move 3 steps in cyclic order to go to point '3)'. So last digit of the first part is 7.
Similarly cycle for 7 for the 2nd part.
1) 7^1= 7 is the last digit.
2) 7^2= 9 is the last digit.
3) 7^3= 3 is the last digit.
4) 7^4= 1 is the last digit.
an then for 7^5 onward's the cycle repeats.
And hence find the last digit in the similar fashion. It comes to 3.
So total is 7+3= 10. So last digit is 0.

Is this solution Helpfull? Yes (1) | No   

deetipriya dasgupta  (10 years ago)

ans is 4. the last digits when 3 is raised to the following powers:
3^1=3
3^2=9
3^3=7
3^4=1
3^5=3
3^6=9
3^7=7
3^8=1. thus cyclicity of 3 is 4. Dividing the power by 4 we get ((1023^4)^980)*1023^3. Thus last digit in this case is 7. Similarly for the second part the last digit obtained s 7. Hence the overall last digit is 4

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Rajat Subhra Bhowmick  (10 years ago)

4
1023^((5*784)+3)=......1
3087^((5*785)+2)=......3
1+3=4

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suparna das  (10 years ago)

23+87=110 so last digits=0

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Guru Pranav  (10 years ago)

laast digit is "0"

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