Latest TCS Aptitude Question SOLUTION: find last digits of (1023^3923)+(3081^3921)

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find last digits of (1023^3923)+(3081^3921)

Asked In TCS Mehendi Chakrabarty (9 years ago)

Solved BHANU KUMAR Read Solution (20)

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BHANU KUMAR  (9 years ago)

we have (1023^3923)+(3081^3921)
(3923/4)=4n+3 (3921/4)=4n+1 where n can be any integer
now through cyclicity we can obtain or you can learn the chart from arun sharma
number ending in 4n+1 4n+2 4n+3 4n
1 1 1 1 1
2 2 4 8 6
3 3 9 7 1


in this way 7+1=8 is the unit's place digit

Is this solution Helpfull? Yes (26) | No (6)  

Aravind  (9 years ago)

A useful tip to solve this type of problems :

power cycle for numbers from 1 to 9
(1) --- {1}
(2) --- {2,4,8,6}
(3) --- {3,9,7,1}
(4) --- {4,6}
(5) --- {5}
(6) --- {6}
(7) --- {7,9,3,1}
(8) --- {8,4,2,6}
(9) --- {9,1}

In this case it's (1023^3923)
From the above table we can see that the number of elements in power cycle for 3 is 4 i.e {3,9,7,1}
So divide 3923 by 4 and note the reminder. Here the reminder is 3 and so 7 will be the answer as 7 is present in the third position of power cycle.

Similarly get the last digit of (3081^3921) as 1 and then add both
1+7=8 is the answer.

Is this solution Helpfull? Yes (20) | No (1)  

Pranoy Bej  (9 years ago)

last digit of 1023^3923 = 7
and last digit of 3081^3921 = 1
so, 7+1 = 8

Is this solution Helpfull? Yes (9) | No (2)  

Sayan Chakraborty  (9 years ago)

1023^3923+3081^3921
dividing both the exponential powers with 4
therefore, 3923/4=4n+3 and 3921/4=4n+1
we know,
4n+1 4n+2 4n+3 4n+4
for 1 1 1 1 1
for 2 2 4 6 8
for 3 3 9 7 1
and so on...
therefore, 4n+3(for 3)=7 and 4n+1(for 1)=1
therefore the last digits will be 7+1 i.e 8

Is this solution Helpfull? Yes (7) | No (1)  

Pooja  (9 years ago)

3^3+1^1=7+1=8

Is this solution Helpfull? Yes (5) | No   

sapna upreti  (9 years ago)

2
1023^3923 as 3^4 will give 3 in unit place thus 3923 must be a divisible by 4
thus 1023^3*1023^3920=7*3=21 thus 1 in unit place
3081^3921=1 in unit place
thus1+1==2

Is this solution Helpfull? Yes (3) | No (6)  

Pallabi Mitra  (9 years ago)

8
unit digit in 1023^3923=unit digit of 3^3923= unit digit in [(3^4)^980*3^3]
= unit digit in [1^980] * unit digit
in [27] = 1 * 7 = 7
unit digit in 3081^3921=unit digit in 1^3921=1
therefore unit digit of (1023^3923)+(3081+3921)=1+7=8
Ans:8

Is this solution Helpfull? Yes (3) | No   

vignesh  (9 years ago)

1023 take unit digit 3 .cyclicity count of 3 is 4 so now divide 3923 by 4 remainder is 3 ,3 9 7 1 is cyclicity level 3rd pace is 7 so last digit of 1023^3923 is 7.
for 3081 its unit digit is 1 .1 cyclicity count is 1 so now divide 3921 by 1 remainder will be 0 ,cyclicity level of 1 is 1 only , for remainder 0 we select last level of cyclicity of 1 so unit digit is 1 .
hence totally 7+1=8
ans is 8.

Is this solution Helpfull? Yes (3) | No (1)  

jyotsna  (9 years ago)

power cycle of 3 is 4.so divide last 2 digits of power value by 4.
23%4=3.so last digit of first expression is last digit of 3^3=27
next expression contains last digit as 1.so 7+1=8

Is this solution Helpfull? Yes (2) | No   

BHANU KUMAR  (9 years ago)

we have (1023^3923)+(3081^3921)
(3923/4)=4n+3 (3921/4)=4n+1 where n can be any integer
now through cyclicity we can obtain or you can learn the chart from arun sharma
number ending in 4n+1 4n+2 4n+3 4n
1 1 1 1 1
2 2 4 8 6
3 3 9 7 1


in this way 7*1=7 is the unit's place digit
in

Is this solution Helpfull? Yes (1) | No (4)  

venkatesh  (9 years ago)

using cyclicity of powers.
(1023^3923) ends with 7 and 3081^3921 ends with 1
so their sum has 8 in its units place.

Is this solution Helpfull? Yes (1) | No   

guddu  (9 years ago)

we have (1023^3923)+(3081^3921)
last digit is 3^3+1^1=27+1=28
last digit is 8

Is this solution Helpfull? Yes (1) | No (2)  

Vijay Pareek  (9 years ago)

0 will be last digit from first we can get 3 as last digit and from last we can get 1 as last digit

Is this solution Helpfull? Yes | No (5)  

sandeep  (9 years ago)

According to power cycle 1023^3923 ends with number 7
and 3081^3921 ends with number 1
then 7+1=8.

Is this solution Helpfull? Yes | No   

LOVENEET SALHOTRA  (9 years ago)

1023 has unit place 3
so 3 raised to 1 is 3
3 raised to 2 is 9
3 to 3 27
3 to 4 81
3 to 5 243
means unit place will change 3 9 7 1 3 9 7 1.......
so now depend on 3923,there is 4 terms after repition occurs
so 3923/4=980 and remainder 3 and at 3rd place 7 is there
now 1 raised to any thing gives 1 at unit place
when they add up gives 7+1=8
ans

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sutapa mukherjee  (9 years ago)

here 3^3923. 3 is in group c so 3923/4 remainder is 3 so 3^3=unit digit 7
next 1^anything+init digit is 1
so (7+1)=8 ans

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Madhur Rusia  (9 years ago)

power of 3 repeats after 3 times and power 1 is always 1.
3923=1307*3 + 2
hence unit place of 1023^3923 = 3^2 =9
unit place of second term is 1
thus unit place in problem is 0 (9+1=10)

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Souvik Das  (9 years ago)

last digits means=unit digit,
now (3^3923)+(1^3921)
=(3)^(last two digit/4=reminder 3)+(1)
=3^3 +1=7+1=8

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RAHUL RANJAN  (9 years ago)

we have (1023^3923)+(3081^3921)
(3923/4)=4n+3 (3921/4)=4n+1 where n can be any integer
now through cyclicity we can obtain or you can learn the chart from arun sharma
number ending in 4n+1 4n+2 4n+3 4n
1 1 1 1 1
2 2 4 8 6
3 3 9 7 1


in this way 7*1=7 is the unit's place digit
in

Is this solution Helpfull? Yes | No (2)  

GAGANJIT SINGH GUJRAL  (9 years ago)

cycle of 3=3,9,7,1. so divide 3923/4. we are left with 3. so 7 is in the 3rd position in the cycle of 3.
cycle of 1=1
so ans is 7+1=8

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