if v w x y z are non negative integer each less than 11 then how many distinct combinations w

So Ans is B) = 1 Only.
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Remaining solution...

Let's analyze if further option available satisfying the given condition. v,w,x,y,z should be less than 11 and greater than zero. The solution we got have v as 10. Suppose v=9. With this, the maximum value possible under the given constraints is

9(11^4)+10(11^3)+10(11^2)+10(11^1)+10=146409
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..
9(11^4)+10(11^3)+10(11^2)+10(11^1)+10=146409
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v(11^4)+w(11^3)+x(11^2)+y(11)+z=151001

151001/11 gives remainder 4.

v(11^4)+w(11^3)+x(11^2)+y(11) = 151001 - z
to make it divisible by 11, z should be 4..
as value of z cannot be any other number, we can form only on 1 distinct combination..
Answer is (B)-1
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ans is b only
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I am sorry to say that your question was wrong. Because there is no mention area, perimeter....
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v(11^4)+w(11^3)+x(11^2)+y(11)+z=151001=11*13727+4
therefore, z=4,next
11*v(11^3)+11*w(11^2)+11*x(11^1)+y(11)=11*13727
11*v(11^2)+11*w(11^1)+11*x+y=13727=11*1247+10
where,y=10,then
11*v(11^1)+11*w+x=1247=11*113+4
where,x=4,then
11*v+w=113=11*10+3
where,w=3,then
11*v=11*10
then,v=10
therefor,z=4,y=10,x=4,w=3,v=10,
ans is 3.
9 years agoHelpfull: Yes(0) No(1)

if v,w,x,y,z are non negative integer, each less than 11,then how many distinct combinations(w,v,x,y,z) satisfy v(11^4)+w(11^3)+x(11^2)+y(11)+z=151001.
optn
a)0
b)1
c)2
d)3