Latest Elitmus Aptitude Question SOLUTION: If v, w,x,y,z are non negative integers, each less than 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..?? a.0 b.1 c.2 d.

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(#M40158267) ELITMUS Numerical Ability question Number System Keep an EYE

If v, w,x,y,z are non negative integers, each less than 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..??

a.0
b.1
c.2
d.3

Asked In Elitmus Shashank Saha (7 years ago)

Unsolved Read Solution (5)

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Read Solution (5) :

Gaurav Kumar gupta  (7 years ago)

Dividing 151001 by 11^4 give quotient 10 and remainder 4591

So

151001 = 10*11^4 + 4591

Dividing 4591 by 11^3 gives quotient 3 and remainder 598

So

151001 = 10*11^4 + 3*11^3 + 598

Dividing 598 by 11^2 gives quotient 4 and remainder 114

So

151001 = 10*11^4 + 3*11^3 + 4*11^2 + 114

Dividing 114 by 11 gives quotient 10 and remainder 4

So

151001 = 10*11^4 + 3*11^3 + 4*11^2 + 10*11 + 4

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

Is this solution Helpfull? Yes (16) | No (1)  

Akeel Mansuri  (7 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

Is this solution Helpfull? Yes (10) | No (1)  

rashmi negi  (7 years ago)

refer this link for ans:
https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.857977.html

Is this solution Helpfull? Yes (5) | No   

rajesh dharamdasani  (7 years ago)

how we are getting quotient 10 and remainder 4591 ?

Is this solution Helpfull? Yes | No (3)  

Mohamad zeaba Thasleem  (1 year ago)

Answer is 1

Is this solution Helpfull? Yes | No (1)  

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