Latest Elitmus Aptitude Question SOLUTION: Q. (999^x -999^y)/1000 what are the max number of remender including 0.x and y can be any natural number

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(#M40031094) ELITMUS Numerical Ability question quant asked on 29 Keep an EYE

Q. (999^x -999^y)/1000
what are the max number of remender including 0.x and y can be any natural number

Asked In Elitmus ravi (10 years ago)

Unsolved Read Solution (7)

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ashes  (10 years ago)

i think 3 is the answer
bcoz acc. to remainder theorem:(-1)^x - (-1)^y
case 1: x and y both even
ans. 0
case 2: x and y both odd
ans. 0
case 3: x-even and y-odd
ans. 2
case 4: x-odd and y-even
ans.-2
case 5: both zero
ans 0

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Vijay Singh Adhikari  (10 years ago)

ans=3
remainders are 0,2,998

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Himanshu upadhyay  (10 years ago)

to be divisible by 1000 we are concerned with last 3 digits,that should be zero
999^x-999^y when x=y then it would generate remainder as 0.
takes x=2,y=1
999*998=997002
take x=4 y=1
999*(999^3 -1 )=997002998*999=996005995002
take x=8,y=1
.i.e 992027944069944027991002
now take x=4,y=2
we get 995008993002
now takes x=6,y=2
we get 994014980013996000
look for last three digits,either its 998,000,or 002
in case of negative no subtract from 1000 like 1000-998=002
therefore answesr is 3,ie 000,002,998

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amrita singh  (9 years ago)

for a^n/(a+1) rem=1 if n is even & rem=-1 if n is odd.
so, for (999^x-999^y)/1000;
if x & y both odd rem= -1-(-1)=0
else if x & y both even rem= 1-1=0
else if x is odd n y is even=-1-1=-2
else if x is even n y is odd=1-(-1)=2
so, 3 rem=0,2,-2
-2 is 998 or (1000-2)

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N Arun Singh  (10 years ago)

@vijay how did u do ? ..i got the ans by takin values ...can u plz explin ?

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pavithra  (10 years ago)

can u please explain it..

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SAMAGRA SHRIVASTAVA  (9 years ago)

9 has a cyclicity of 3 hence 9^n will give either 9 or 1 at units place....
hence there can be ***9-***1 where divident is having 8 as unit digit
or it can be ***1-***9 where divident is having (11-9) 2 as unit digit
or it can be ***9-***9 where divident is having 0 as unit digit
hence ans will be 3.

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