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

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(#M40040029) ELITMUS Numerical Ability question Keep an EYE

If v,w,x,y and z are non negative integers, each less than 11, than how many distinct combination of (v,w,x,y,z) satisfy v(11^4) + w(11^3) + x(11^2) + y(11) + z=151001

Asked In Elitmus Ravi Bhardwaj (9 years ago)

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Gopikrishnamaraju  (9 years ago)

11(11^3v+11^2w+11x+y) + z = 11x13727 + 4 [hence; z = 4]
repeating the above procedure;
11(11^2v + 11w +x) + y =11x1247 + 10 [y = 10]
11(11v + w ) + x = 11x113 + 4 [x = 4]
11v + w = 11x10 + 3 [v=10 , w=3]
So one combination

Is this solution Helpfull? Yes (41) | No (1)  

Yousuf Waqar  (9 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

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Narasapu Hari prasad  (9 years ago)

only one combinatin satisfy
the value of v=10
w=3
x=4
y=10
z=4
which is in base 11 form.

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himannshu kumar  (9 years ago)

bhai log maine elitmus ka exam 28 september ko diya tha kya hum phir se 5 oct.ko de sakte hai same id se ,or dono ka result declare hoga ya jo last me diye uska pls batao

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Ravi Bhardwaj  (9 years ago)

Please explain how did you recognize Hari?

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Narasapu Hari prasad  (9 years ago)

@Ravi Bhardwaj
formula and v,w,x,y,z

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Narasapu Hari prasad  (9 years ago)

@Ravi Bhardwaj
formula and constraints v,w,x,y,z

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pawan   (9 years ago)

36 by basics algebra identities frst by subtacting 2 we can form it square and then soving

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