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
options A)0 B) 1 C)2 D)3

• v(11^4)+w(11^3)+x(11^2)+y(11)+z=151001
  = > v(11^4)+w(11^3)+x(11^2)+y(11)=151001-z
  Looking at the equation we can get that LHS is always divisible by 11,so for (151001-z)to be divisible by 11,z value should be only 4.Answer is 1
• 11 years agoHelpfull: Yes(17) No(1)
• v=10 w=3 x=4 y=10 z=4
  only one combination
  
• 11 years agoHelpfull: Yes(1) No(2)
• there is a no. base 11
  so when we convert no. 151001 into base 11 we get,
  v=10,w=3,x=4,y=10,z=4
  therefore only 1 combination ie (10,3,4,10,4)
• 11 years agoHelpfull: Yes(1) No(0)
• 1 i got one comb i.e., v=1,w=0,x=3,y=8,z=2
• 11 years agoHelpfull: Yes(0) No(7)
• m still not sure about answer
• 11 years agoHelpfull: Yes(0) No(0)
• See the URL
  https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.857977.html
• 8 years agoHelpfull: Yes(0) No(0)