Latest Elitmus Aptitude Question SOLUTION: If v,w,x,y, and 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)

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(#M40155158) ELITMUS Numerical Ability question number systems Keep an EYE

If v,w,x,y, and 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 Jagadeesh Ginjupalli (8 years ago)

Unsolved Read Solution (3)

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Read Solution (3) :

Paplesh k. gour  (8 years ago)

Taking 11 common from LHS separating z and breaking 151001 as a factor of 11 with some remainder in RHS
11(113v+112w+11x+y) + z = 11x13727 + 4 [hence; z = 4]
Equating z =4 and cancelling 11 from each side we get the following simplified equation and again repeating the above procedure;
11(112v + 11w +x) + y =11x1247 + 10 [y = 10]
11(11v + w ) + x = 11x113 + 4 [x = 4]
11v + w = 11x10 + 3 [v=10 , w=3]
So we get the unique solution for the above equaton.

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dhwani sondhi  (8 years ago)

b) 1
v=10, w=3, x=4, y=10,z=4
is the only solution

Is this solution Helpfull? Yes (3) | No (2)  

NIDHISH  (7 years ago)

11(113v+112w+11x+y) + z = 11x13727 + 4 [hence; z = 4]
Equating z =4 and cancelling 11 from each side we get the following simplified equation and again repeating the above procedure;
11(112v + 11w +x) + y =11x1247 + 10 [y = 10]
11(11v + w ) + x = 11x113 + 4 [x = 4]
11v + w = 11x10 + 3 [v=10 , w=3]
So we get the unique solution for the above equaton.

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