Latest TCS Aptitude Question SOLUTION: If x^y denotes x raised to the power of y find last two digits of (1507^3381)+(1437^3757)

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Number System

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40152704) TCS Numerical Ability question number tricks Keep an EYE

If x^y denotes x raised to the power of y find last two digits of (1507^3381)+(1437^3757)

Asked In TCS Debasmit Guchhait (8 years ago)

Unsolved Read Solution (12)

Is this Puzzle helpful?   (4)   (2) Submit Your Solution Number System

Read Solution (12) :

milan jena  (8 years ago)

soln is 44
bcoz power cycle of 7 is 4.when 3381 is divided by 4 it gives remainder 1, so (07)^1=07
similarly in 3757 it is (37)^1=37
07+37 =44
but tcs question was(1457^3757)

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

Mahesh Malviya  (8 years ago)

solution is 44
because we know that (07)^4 gives unit digit is 1 . so we can write power 3381 in the multiplication of power 4. now 3381 divide by 4 than remainder is 1 , so (07)^1 =07.

similarly (37)^3757 gives the remainder 1 when 3757 is divided by 4 . so (37)^1=37.

now 07 + 37 = 44

Is this solution Helpfull? Yes (4) | No   

Brajeshrai  (8 years ago)

07+61^939*37=07+17=24

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

Golluru Lalitha  (8 years ago)

7^1=7 (last digits )
7^2=9
7^3=3
7^4=1 further same series will repeats. convert the eqn into powers of 7^4 then
(7^4)^845*7 + (7^4)^939*7
last digits = 1*7 + 1*7 =14
.". last digit =4

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

shradha dhondiram mirkale  (8 years ago)

We have ,7^(multiple of 4)=1
So let us consider last two digits of 3381 i.e. 81 and divide 81 by 4 so that we will get 40
Quotient and remainer as 1,so 7^4*40=1 and as remainder is 1 so 7^1=7
Now consider second no.=3757 take its last two digits=57
Divide by 4 wll get quotient=14,so 7^4*14=1
And remainder is =1,so 7^1=7
So, in both cases we have (7^1)+(7^1)=14
Hence and.=14

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

Rajanikanta Pradhan  (8 years ago)

solution is 24

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

utkarsh singh  (8 years ago)

4
Since last digit is 7 in both the nos. ,we can observe easily that 7 raised till power 4 repeat itself .

Is this solution Helpfull? Yes | No (1)  

goutham  (8 years ago)

78 (49+29)

Is this solution Helpfull? Yes | No (2)  

sonu kumar  (8 years ago)

(1507^3381)+(1437^3757)=76

Is this solution Helpfull? Yes | No (1)  

anusha  (8 years ago)

is it asked again?

Is this solution Helpfull? Yes | No (2)  

Amanulla Khan  (8 years ago)

07+17 = 24

Is this solution Helpfull? Yes | No (1)  

bibhudatta  (8 years ago)

First for cyclicity of 7 is 4 ,for a unit place odd number the whatever the power it should give 01,for another one we should go for calculation so 1507^3381 can be written as 1507^(4*845)+1so for this the last digit for 1507^4*835 is always 01 and for 1507^1 it is 07 so last 2 digit is 01*07=7..similarly for 1437^(4*939)+1 is 01*37 so 37 and the sum of last 2 digit will be 07+37=44...

Is this solution Helpfull? Yes | No (1)  

Please Login to Read Solution.