Latest Elitmus Aptitude Question SOLUTION: v,w,x,y,z are non negative integers each <11, then how many distinct combinations of (v,w,x,y,z) satisfy v(11^4)+w(11^3)+x(11^2)+y(11)+z= 151001

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(#M40034312) ELITMUS Numerical Ability question plynomials Keep an EYE

v,w,x,y,z are non negative integers each <11, then how many distinct combinations of (v,w,x,y,z) satisfy
v(11^4)+w(11^3)+x(11^2)+y(11)+z= 151001

Asked In Elitmus Geetanshu (10 years ago)

Solved RAKESH Read Solution (7)

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Read Solution (7) :

RAKESH  (10 years ago)

v(11^4)+w(11^3)+x(11^2)+y(11)+z= 151001

=> 11*( v*11^3 + w*11^2+ x*11 + y ) + z = 11*13727 + 4

=> ( v*11^3 + w*11^2+ x*11 + y )= 13727 & z = 4

=> 11*( v*11^2 + w*11 + x ) + y = 11*1247 + 10

=> ( v*11^2 + w*11 + x )= 1247 & y = 10

=> 11*( v*11 + w ) + x = 11*113 + 4

=> ( v*11 + w )= 113 & x = 4

=> ( v*11 + w ) = 11*10 + 3

=> v = 10 & w = 3

so, v=10, w=3, x=4, y=10 ,z=4

(v,w,x,y,z)=(10,3,4,10,4)

so, only one combination

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Geetanshu  (10 years ago)

options were>> 0,1,2 or 3

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tausif  (10 years ago)

Answer is 1
because there is one particular solution for this

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saurabh agarwal  (10 years ago)

ans is 0.
because when we solve we get v=10,w=3,x=4,y=10,z=4.
here all are not distinct.

Is this solution Helpfull? Yes (3) | No (15)  

SANTANA PANDA  (10 years ago)

v(11^4)+w(11^3)+x(11^2)+y(11)+z= 151001=>
11*( v*11^3 + w*11^2+ x*11 + y ) + z = 11*13727 + 4
=> ( v*11^3 + w*11^2+ x*11 + y )= 13727 & z = 4
=> 11*( v*11^2 + w*11 + x ) + y = 11*1247 + 10

=> ( v*11^2 + w*11 + x )= 1247 & y = 10
=> 11*( v*11 + w ) + x = 11*113 + 4
=> ( v*11 + w )= 113 & x = 4
=> ( v*11 + w ) = 11*10 + 3
=> v = 10 & w = 3so, v=10, w=3, x=4, y=10 ,z=4(v,w,x,y,z)=(10,3,4,10,4) so, only one combination.

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Nishant Anand  (10 years ago)

Rakesh z=5 in first line

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naga malleswara rao yaganti  (9 years ago)

@Rakesh z can be any number of form 11n+4 ?

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