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(#M40061211) ELITMUS Numerical Ability question number system Keep an EYE

Find the sum of its last five digits (2020202) ^4

Asked In Elitmus ratnesh (10 years ago)

Unsolved Read Solution (11)

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Read Solution (11) :

udaykiran  (10 years ago)

this is equal to 2^4*(1010100+1)^4.

now use binomial.

2^4*[(1010100)^4+4*(1010100)^3+6*(1010100)^2+4*(1010100)+1].

so last 5 digits=16[6*10000+40400+1]=16*00401=06416.

so sum=17

Is this solution Helpfull? Yes (36) | No (4)  

neeraj kumar  (10 years ago)

the answer will be 17

Is this solution Helpfull? Yes (3) | No   

arpit farkya  (10 years ago)

Right ans is 18 =16*0405

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

arpit farkya  (10 years ago)

7=00016
Thats it

Is this solution Helpfull? Yes | No (7)  

arpit farkya  (10 years ago)

Right ans is 18 =16*0405

Is this solution Helpfull? Yes | No (7)  

vinay jain  (10 years ago)

answer should be 36

Is this solution Helpfull? Yes | No (3)  

ghanshyam gupta  (10 years ago)

answer is 17

Is this solution Helpfull? Yes | No   

suresh  (9 years ago)

Answer is 17 and it is 100% sure

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swarup  (9 years ago)

ans is 23 not 17

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shilpa  (8 years ago)

ans is 22 bcoz 2020202^2=last five digits=20804*2020202=last five digits=82408
so sum is 8+2+4+0+8=22

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VIVEK RAWAT  (8 years ago)

let's multiply only last 5 digit (20202)^4=(20001+201)^4
=((20001+201)^2)^2
=(00001+40001+2*20001*201)^2) (a+b)^2 considering only last five digit
=(00001+40001+40402)^2
=(80404)^2
= 03216 =12ans; or use (20001+201)^4 through binomial expansion

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